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L-17: Electric Current

Physics — Class 12 · NIOS Code 312 · Module 5 · Source: 312_Physics_Eng_Lesson17.pdf

Electric Current

Electricity powers lamps, fans, TVs, and countless devices. While earlier lessons dealt with charges at rest, this lesson studies charges in motion — how current depends on potential difference, how circuits are analysed, and how emf, resistance, and power are measured.

NIOS objectives: Ohm's law; ohmic vs non-ohmic resistance; series/parallel resistors; primary vs secondary cells; Kirchhoff's rules; Wheatstone bridge; potentiometer for emf and internal resistance.

Free and Bound Electrons

Atoms are normally neutral. Valence electrons in metals are loosely bound and become free when a small potential difference is applied, enabling conduction.

17.1 Electric Current

When potential difference is applied across a conductor, free electrons drift opposite to the field — constituting current. Conventionally, current direction is that in which a positive charge would move (opposite to electron flow).

I = Δq / Δt  |  I = dq/dt
I = current (A = ampere)
Δq = charge crossing a surface normal to flow (C)
1 A = 1 coulomb per second
Smaller units: mA (10⁻³ A), μA (10⁻⁶ A)

Current may arise from electrons (metals), holes and electrons (semiconductors), or ions (electrolytes).

Fig 17.1 — Charge Flow Through Conductor area A − electron drift (opposite to I conv.) conventional current I →
Fig 17.1 — Current = rate of charge transfer across surface A perpendicular to flow
I = n A e vd
n = free electron density (m⁻³)
A = cross-sectional area (m²)
e = electron charge (1.6×10⁻¹⁹ C)
vd = drift velocity (m·s⁻¹) — see §17.9

17.2 Ohm's Law

Ohm (1828): current through a conductor is directly proportional to potential difference across it, provided temperature, pressure, and other physical conditions remain constant.

V = R I  |  R = V / I
R = resistance (Ω = ohm)
1 Ω = 1 V/A
I–V graph is a straight line through origin for ohmic conductors
Metals obey Ohm's law (linear region)
  • Ohmic: linear V–I relation (most metals, electrolytes under certain conditions).
  • Non-ohmic: vacuum diode, semiconductor diode, transistors — nonlinear I–V curve.

17.2.1 Resistance and Resistivity

Resistance depends on length l and area A:

R = ρ l / A
ρ = resistivity / specific resistance (Ω·m)
R ∝ l; R ∝ 1/A
Resistivity depends on material, not dimensions
Conductivity: σ = 1/ρ (S·m⁻¹ or mho·m⁻¹)

If l = 1 m and A = 1 m², then ρ = R. Doubling wire length doubles R; doubling diameter (4× area) quarters R.

17.3 Grouping of Resistors

Equivalent resistance is a single resistance allowing the same current as the combination for the same applied voltage.

17.3.1 Series Combination

R = R₁ + R₂ + R₃ + …
Same current I through each resistor
Potential differences add: V = V₁ + V₂ + …
Used to reduce voltage across a component (e.g. lamp)

17.3.2 Parallel Combination

1/R = 1/R₁ + 1/R₂ + 1/R₃ + …
Same potential V across each branch
Currents add: I = I₁ + I₂ + …
Req < smallest individual R
Home appliances (bulbs, fans) wired in parallel at 220 V
Fig 17.8 & 17.9 — Series and Parallel Resistors Series — same I R₁ R₂ R = R₁ + R₂ Parallel — same V R₁ R₂ 1/R = 1/R₁ + 1/R₂ Home mains 220 V Bulb ∥ Fan ∥ Heater Each with own switch More appliances → lower R_eq, higher I B F H
Fig 17.8–17.10 — Series adds R; parallel divides current; home appliances in parallel

17.4 Types of Resistors

Wire-wound: manganin, constantan, or nichrome wire on insulating cylinder. Carbon: molded carbon cylinder with wire leads.

Colour code: R = AB × 10C Ω ± D% (first two digits A, B; third = multiplier; fourth = tolerance).

  • Gold tolerance = 5%; Silver = 10%; Body colour = 20%.
  • Example: Blue-Grey-Green-Silver → 68 × 10⁵ Ω ± 10% = 6.8 MΩ ± 10%.

17.5 Temperature Dependence of Resistance

ρ = ρ₀ [1 + α (T − T₀)]
α = temperature coefficient of resistivity (°C⁻¹)
Most metals: ρ increases with T (linear over limited range)
R = R₀[1 + α(T − T₀)]
  • Superconductors: zero resistivity below transition temperature.
  • Alloys (manganin, constantan, nichrome): α ≈ 10⁻⁶ °C⁻¹ — used for standard resistances.
  • Semiconductors: ρ usually decreases with T (negative α).

17.6 EMF and Potential Difference

EMF (E) of a cell equals the potential difference between terminals when no current is drawn (open circuit). When current I flows, internal resistance r causes a drop Ir:

E = V + I r
E = electromotive force (V)
V = terminal potential difference when current flows
r = internal resistance of cell
EMF depends on electrolyte, electrodes, temperature — not on cell size

17.6.1 Primary and Secondary Cells

  • Primary cells: chemical energy → electrical energy directly; consumed and not rechargeable (dry cell, Daniel cell, Voltaic cell).
  • Secondary cells: reversible reaction; can be recharged (lead-acid accumulator in cars/inverters).

17.7 Kirchhoff's Rules

For complex networks beyond simple Ohm's law analysis.

(i) Junction Rule (First Rule)

Sum of currents directed toward a junction equals sum directed away — no charge accumulation at a point in steady state.

Σ I (toward junction) = Σ I (away from junction)
Algebraic sum of all currents at a junction = 0
Extension of charge continuity in circuits
Fig 17.16 — Kirchhoff Junction Rule A I₁ I₂ I₃ I₁ + I₂ = I₃ (at junction A)
Fig 17.16 — Currents into junction A equal currents leaving

(ii) Loop Rule (Second Rule)

Algebraic sum of potential differences around any closed loop is zero — conservation of energy.

Σ (I R) = Σ E  |  Σ ΔV = 0
Traverse loop clockwise or anticlockwise
IR drop positive in direction of current
EMF positive from − to + through cell
Apply to each independent mesh in the network

17.7.1 Wheatstone Bridge

Four resistances P, Q, R, S in a bridge network. Galvanometer G detects null when B and D are at same potential:

P / Q = R / S  →  S = Q R / P
Balance condition independent of battery emf
Galvanometer needs no calibration (null method)
Best sensitivity when all arm resistances are nearly equal
Accurate for low resistances
Fig 17.19 — Wheatstone Bridge A C B D G (null when balanced) P Q R S E, K₁ Balance: P/Q = R/S · S unknown measured
Fig 17.19 — Null galvanometer when P/Q = R/S

17.8 Potentiometer

A versatile instrument using the null method — draws no current from the source being measured. Uniform wire AB (often 10 wires in series) with jockey; potential drops linearly along wire.

V = (E / l) × l₁ = k l₁
E = emf of driving cell (must exceed unknown V)
l = total wire length; l₁ = balance length
At null point, galvanometer shows zero deflection

17.8.3 Comparison of EMFs

E₁ / E₂ = l₁ / l₂
Same potentiometer wire and driver cell
Balance lengths l₁, l₂ for cells E₁, E₂ respectively

17.8.4 Internal Resistance of a Cell

With resistance R shunted across cell (key closed), terminal voltage V₁ < E₁. Balance lengths l₁ (open) and l₂ (closed):

r = (l₁/l₂ − 1) × R
r = internal resistance
E₁/V₁ = l₁/l₂ and V₁ = E₁R/(R+r)
r depends on plate area, separation, electrolyte strength

17.9 Drift Velocity of Electrons

Conduction electrons move randomly at ~10⁶ m/s. With no field, average velocity is zero. Under field E, electrons drift slowly (~10⁻⁴ m/s) opposite to E, losing excess energy in collisions (heating the conductor).

vd = (e E τ) / m
τ = average time between collisions (s)
m = electron mass
Combining with I = nAe vd gives Ohm's law
ρ = m / (n e² τ)
Microscopic origin of resistivity
More collisions (smaller τ) → higher ρ
Derives R = ρl/A from electron dynamics

17.10 Power Consumed in an Electrical Circuit

P = V I = I² R = V² / R
P = power (watt, W)
Rate of energy dissipation as heat (Joule heating)
Joule's law: Q = I² R t
1 kWh = 1 unit of domestic electricity
         ELECTRIC CURRENT — KEY FORMULAS
         ================================
    Current          :  I = dq/dt = nAe v_d
    Ohm's law        :  V = IR
    Resistivity      :  R = ρl/A ;  σ = 1/ρ
    Series R         :  R = R₁ + R₂ + …
    Parallel R       :  1/R = 1/R₁ + 1/R₂ + …
    Cell             :  E = V + Ir
    Kirchhoff loop   :  ΣIR = ΣE
    Wheatstone       :  P/Q = R/S
    Potentiometer    :  E₁/E₂ = l₁/l₂
    Power            :  P = VI = I²R = V²/R

Quick Revision

  • Current: rate of charge flow; 1 A = 1 C/s; conventional direction opposite to electrons.
  • Ohm's law: V = IR; ohmic (metals) vs non-ohmic (diodes).
  • ρ = RA/l; series R adds; parallel reciprocals add.
  • EMF E = V + Ir; primary (disposable) vs secondary (rechargeable) cells.
  • Kirchhoff: junction (charge conservation); loop (energy conservation).
  • Wheatstone: P/Q = R/S; potentiometer: null method, no current drawn.
  • Drift velocity vd = eEτ/m; power P = I²R.
20 cards · click any card to flip
Electric current
Rate of transfer of charge across a surface normal to flow. I = dq/dt. Unit: ampere (1 A = 1 C/s).
Conventional current direction
Direction in which positive charge would move — opposite to electron drift in metals.
Ohm's law
V ∝ I when temperature and physical conditions are constant. V = IR. Metals are ohmic (linear V–I).
Resistance
Property opposing current flow. R = V/I. Unit: ohm (Ω). 1 Ω = 1 V/A.
Resistivity
ρ = RA/l (Ω·m). Material property — resistance of 1 m wire with 1 m² cross-section. σ = 1/ρ.
Ohmic vs non-ohmic
Ohmic: linear V–I (metals). Non-ohmic: diodes, transistors — nonlinear I–V curves.
Series resistors
R = R₁ + R₂ + R₃ + … Same current through each; voltages add.
Parallel resistors
1/R = 1/R₁ + 1/R₂ + … Same voltage across each; currents add. R_eq < smallest R.
Temperature and resistance
ρ = ρ₀[1 + α(T−T₀)]. Metals: ρ increases with T. Semiconductors: ρ usually decreases with T.
EMF vs terminal voltage
EMF E = open-circuit terminal voltage. With current: E = V + Ir. V drops as I increases.
Primary vs secondary cells
Primary: not rechargeable (dry cell). Secondary: reversible reaction, rechargeable (lead-acid battery).
Kirchhoff junction rule
Sum of currents into a junction = sum leaving. No charge accumulation in steady state.
Kirchhoff loop rule
Σ(IR) = ΣE around any closed loop. Algebraic sum of potential differences = 0 (energy conservation).
Wheatstone bridge balance
P/Q = R/S. Galvanometer null. S = QR/P. Independent of driver emf; null method needs no galvanometer calibration.
Potentiometer principle
Potential drops linearly along uniform wire. V = kl₁ at balance. Draws no current from measured source (null method).
Compare emfs (potentiometer)
E₁/E₂ = l₁/l₂ using same wire and driver cell. Balance lengths l₁, l₂ for each cell.
Internal resistance (potentiometer)
r = (l₁/l₂ − 1)R. l₁ with cell alone; l₂ with shunt resistance R across cell.
Drift velocity
v_d = eEτ/m ≈ 10⁻⁴ m/s (much less than random thermal speed ~10⁶ m/s). I = nAe v_d.
Electrical power
P = VI = I²R = V²/R (watt). Joule heating: Q = I²Rt. Home energy in kWh (units).
Resistor colour code
R = AB × 10^C Ω ± tolerance. First two colours = digits; third = multiplier; fourth = tolerance (gold 5%, silver 10%).

Q1. The SI unit of electric current is:

Q2. According to Ohm's law, for an ohmic conductor:

Q3. The resistivity of a material is given by:

Q4. Three resistors 2 Ω, 3 Ω and 6 Ω are connected in parallel. Equivalent resistance is:

Q5. When current is drawn from a cell, the terminal voltage V and emf E are related by:

Q6. Kirchhoff's junction rule is based on conservation of:

Q7. A Wheatstone bridge is balanced when:

Q8. A potentiometer measures emf by:

Q9. Power dissipated in a resistor R carrying current I is:

Q10. Drift velocity of electrons in a conductor under electric field is typically:

I = Δq/Δt
I = n A e v_d
V = R I  |  R = V/I
R = ρl/A
R = R₁ + R₂ + R₃ + …
1/R = 1/R₁ + 1/R₂ + …
ρ = ρ₀[1 + α(T − T₀)]
E = V + I r
Σ I_in = Σ I_out
Σ (I R) = Σ E
P/Q = R/S
V = (E/l) × l₁
E₁/E₂ = l₁/l₂
r = (l₁/l₂ − 1) × R
v_d = e E τ / m
ρ = m/(n e² τ)
P = V I = I² R = V²/R

1. Formulas & Definitions

Full Ch 17 study guide — current, Ohm's law, resistor networks, cells, Kirchhoff, bridge, potentiometer, drift velocity, and power.

I = Δq/Δt

Definition: Electric current — rate of charge transfer across a surface normal to flow.

Derivation

Charge Δq crosses area A in time Δt; instantaneous I = dq/dt.

Variables

I (A) · Δq (C) · Δt (s) · 1 A = 1 C/s

Why it works

Fundamental definition — all circuit analysis starts here.

Historical context

Conventional current = direction positive charge would move (opposite to electron drift).

Deep understanding

Current may be due to electrons (metals), ions (electrolytes), or holes (semiconductors).

2. Diagrams & Visuals

I = Δq/Δt

Color-coded visual · step-by-step breakdown below

  1. Count charge Δq crossing surface
  2. Measure time interval Δt
  3. I = Δq/Δt
  4. Use dq/dt for instantaneous current

3. Solved Examples

Basic

Q: 2 C in 0.5 s.

Solution: I=4 A

Answer: 4 A

Intermediate

Q: 500 mA in mA?

Solution: 0.5 A

Answer: 500 mA

Advanced

Q: Steady 1 A for 10 min.

Solution: q=600 C

Answer: 600 C

Exam

Q: 1 ampere definition?

Solution: 1 C/s

Answer: Sec 17.1

I = n A e v_d

Definition: Microscopic expression for current in a metallic conductor.

Derivation

n electrons per m³ each drift with v_d; charge e per electron; area A.

Variables

n (m⁻³) · A (m²) · e = 1.6×10⁻¹⁹ C · v_d (m/s)

Why it works

Links macroscopic I to electron drift — basis of microscopic Ohm's law.

Historical context

v_d ~ 10⁻⁴ m/s << thermal speed ~10⁶ m/s.

Deep understanding

More free electrons (n), larger wire (A), or faster drift → larger current.

2. Diagrams & Visuals

I = nAe v_d

Color-coded visual · step-by-step breakdown below

  1. Find n, A, v_d
  2. Multiply n×A×e×v_d
  3. Result in amperes

3. Solved Examples

Basic

Q: Double v_d, rest same?

Solution: I doubles

Answer: 2× I

Intermediate

Q: Wire area halved?

Solution: I halves

Answer: Half I

Advanced

Q: n=10²⁸, A=10⁻⁶, v_d=10⁻⁴.

Solution: Order ~1.6 A

Answer: ~1.6 A

Exam

Q: Microscopic current?

Solution: nAe v_d

Answer: Sec 17.1

V = R I  |  R = V/I

Definition: Ohm's law — current proportional to potential difference at constant temperature.

Derivation

Empirical linear relation; slope of I–V graph = 1/R.

Variables

V (V) · I (A) · R (Ω) · 1 Ω = 1 V/A

Why it works

Core relation for resistor networks, power, and meter readings.

Historical context

Georg Ohm (1828); valid for ohmic conductors (most metals).

Deep understanding

Non-ohmic: diodes, transistors — nonlinear I–V curves.

2. Diagrams & Visuals

V = IR Linear I–V for metals

Color-coded visual · step-by-step breakdown below

  1. Identify V and I (or R)
  2. V = IR or R = V/I
  3. Check ohmic conditions
  4. Units: V, A, Ω

3. Solved Examples

Basic

Q: R=10 Ω, I=2 A.

Solution: V=20 V

Answer: 20 V

Intermediate

Q: V=12 V, R=4 Ω.

Solution: I=3 A

Answer: 3 A

Advanced

Q: Double V, R fixed?

Solution: I doubles

Answer: Linear

Exam

Q: Ohm's law?

Solution: V=IR

Answer: Sec 17.2

R = ρl/A

Definition: Resistance of uniform conductor in terms of resistivity ρ, length l, area A.

Derivation

R ∝ l and R ∝ 1/A; ρ is material property.

Variables

ρ (Ω·m) · l (m) · A (m²) · σ = 1/ρ (S/m)

Why it works

Design wires, heating elements, and predict R from geometry.

Historical context

If l=1 m, A=1 m² then R=ρ. Doubling length doubles R; doubling diameter quarters R.

Deep understanding

ρ depends on material and temperature, not wire shape.

2. Diagrams & Visuals

l R = ρl/A

Color-coded visual · step-by-step breakdown below

  1. Length l, area A
  2. Resistivity ρ of material
  3. R = ρl/A
  4. Or ρ = RA/l

3. Solved Examples

Basic

Q: ρ=1.7×10⁻⁸, l=2 m, A=10⁻⁶.

Solution: R≈34 mΩ

Answer: ~0.034 Ω

Intermediate

Q: Length doubled?

Solution: R doubles

Answer: 2R

Advanced

Q: Area quadrupled?

Solution: R quarters

Answer: R/4

Exam

Q: Resistivity formula?

Solution: R=ρl/A

Answer: Sec 17.2.1

R = R₁ + R₂ + R₃ + …

Definition: Equivalent resistance for resistors in series.

Derivation

Same current I; voltages add V = V₁+V₂+… → IR = IR₁+IR₂+…

Variables

R_eq > any individual Rᵢ

Why it works

Voltage dividers, lamps in series, increasing total resistance.

Historical context

Opposite to capacitors — resistors in series add directly.

Deep understanding

Largest R carries same I but largest V drop.

2. Diagrams & Visuals

R = R₁+R₂+…

Color-coded visual · step-by-step breakdown below

  1. Confirm series — same I
  2. Add resistances
  3. R_eq = R₁+R₂+…

3. Solved Examples

Basic

Q: R₁=3 Ω, R₂=7 Ω series.

Solution: R=10 Ω

Answer: 10 Ω

Intermediate

Q: Three 6 Ω series?

Solution: R=18 Ω

Answer: 18 Ω

Advanced

Q: Same V, series vs one R?

Solution: I smaller in series

Answer: Higher R_eq

Exam

Q: Series resistors?

Solution: Add R

Answer: Sec 17.3.1

1/R = 1/R₁ + 1/R₂ + …

Definition: Equivalent resistance for resistors in parallel.

Derivation

Same V across each; currents add I = I₁+I₂+… → V/R = V/R₁+V/R₂+…

Variables

R_eq < smallest Rᵢ

Why it works

Home wiring — appliances at 220 V in parallel; more branches → lower R_eq.

Historical context

Mathematically like capacitors in series (reciprocal sum).

Deep understanding

Smallest R draws largest branch current.

2. Diagrams & Visuals

1/R = Σ1/Rᵢ

Color-coded visual · step-by-step breakdown below

  1. Confirm parallel — same V
  2. Add reciprocals
  3. 1/R_eq = 1/R₁+1/R₂+…
  4. Invert for R_eq

3. Solved Examples

Basic

Q: R₁=R₂=6 Ω parallel.

Solution: R_eq=3 Ω

Answer: 3 Ω

Intermediate

Q: Two 4 Ω parallel.

Solution: R_eq=2 Ω

Answer: 2 Ω

Advanced

Q: R₁=2, R₂=6 Ω parallel.

Solution: R_eq=1.5 Ω

Answer: 1.5 Ω

Exam

Q: Parallel resistors?

Solution: Reciprocal sum

Answer: Sec 17.3.2

ρ = ρ₀[1 + α(T − T₀)]

Definition: Temperature dependence of resistivity (linear approximation).

Derivation

Most metals: more lattice vibrations → more collisions → higher ρ with T.

Variables

α = temp coefficient (°C⁻¹) · T₀ reference temp

Why it works

Explains why filament resistance cold < hot; alloy standards use low α.

Historical context

Also R = R₀[1+α(T−T₀)]. Semiconductors often have negative α.

Deep understanding

Superconductors: ρ→0 below T_c. Manganin/constantan: α ≈ 10⁻⁶ °C⁻¹.

2. Diagrams & Visuals

ρ increases with T (metals)

Color-coded visual · step-by-step breakdown below

  1. Reference ρ₀ at T₀
  2. Find α for material
  3. ρ = ρ₀[1+α(T−T₀)]
  4. Compute R if needed

3. Solved Examples

Basic

Q: α=0.004, ΔT=50°C.

Solution: ρ increases 20%

Answer: 1.2ρ₀

Intermediate

Q: T drops, metal wire?

Solution: R decreases

Answer: Lower ρ

Advanced

Q: Semiconductor heated?

Solution: ρ usually falls

Answer: Negative α

Exam

Q: Temp coefficient?

Solution: ρ=ρ₀[1+αΔT]

Answer: Sec 17.5

E = V + I r

Definition: EMF of cell equals terminal voltage plus internal drop Ir.

Derivation

When current I flows, internal resistance r causes lost potential inside cell.

Variables

E (V) · V = terminal PD · r = internal resistance (Ω)

Why it works

Real cells deviate from ideal; explains why V < E under load.

Historical context

Open circuit I=0 → V=E. Short circuit V≈0, I≈E/r.

Deep understanding

EMF depends on chemistry/temperature, not cell size; r depends on plates, electrolyte.

2. Diagrams & Visuals

E, r V Ir E = V + Ir

Color-coded visual · step-by-step breakdown below

  1. Measure terminal V with current I
  2. Know internal r
  3. E = V + Ir
  4. Open circuit: E = V_oc

3. Solved Examples

Basic

Q: E=1.5 V, I=0.5 A, r=0.2 Ω.

Solution: V=1.4 V

Answer: 1.4 V

Intermediate

Q: No current drawn?

Solution: V=E

Answer: Open circuit

Advanced

Q: r=0 ideal cell?

Solution: V=E always

Answer: Ideal

Exam

Q: Cell terminal voltage?

Solution: E=V+Ir

Answer: Sec 17.6

Σ I_in = Σ I_out

Definition: Kirchhoff junction rule — charge conservation at a node.

Derivation

Steady state: no charge piles up at junction; inflow equals outflow.

Variables

Sign convention: assign direction to each branch current

Why it works

First tool for multi-loop circuits where simple series/parallel fails.

Historical context

Also written: algebraic sum of currents at junction = 0.

Deep understanding

At junction A: I₁+I₂ = I₃. Count independent equations = junctions − 1.

2. Diagrams & Visuals

I₁+I₂ = I₃

Color-coded visual · step-by-step breakdown below

  1. Label junction
  2. Mark currents toward/away
  3. Sum in = sum out
  4. Solve with loop equations

3. Solved Examples

Basic

Q: I₁=2 A, I₂=3 A into junction.

Solution: I_out=5 A

Answer: 5 A

Intermediate

Q: Three branches, two known?

Solution: Find third

Answer: Conservation

Advanced

Q: Steady DC required?

Solution: Yes

Answer: No accumulation

Exam

Q: Junction rule basis?

Solution: Charge conservation

Answer: Sec 17.7

Σ (I R) = Σ E

Definition: Kirchhoff loop rule — algebraic sum of potential differences in closed loop is zero.

Derivation

Conservation of energy: net work per charge around loop = 0.

Variables

IR drop + in direction of I; EMF + from − to + through cell

Why it works

Gives remaining equations to solve unknown currents and emfs.

Historical context

Apply to each independent mesh; consistent sign convention essential.

Deep understanding

Equivalent form ΣΔV = 0. Traverse loop clockwise or anticlockwise consistently.

2. Diagrams & Visuals

ΣIR = ΣE

Color-coded visual · step-by-step breakdown below

  1. Choose closed loop
  2. Assign traversal direction
  3. Sum IR drops and EMFs with signs
  4. Set sum = 0

3. Solved Examples

Basic

Q: One cell E, one R, loop.

Solution: IR = E

Answer: Simple circuit

Intermediate

Q: Two cells opposing?

Solution: Net EMF in sum

Answer: Sign matters

Advanced

Q: Independent loops count?

Solution: Meshes in network

Answer: KVL + KCL

Exam

Q: Loop rule basis?

Solution: Energy conservation

Answer: Sec 17.7

P/Q = R/S

Definition: Wheatstone bridge balance condition — galvanometer shows zero deflection.

Derivation

At balance, B and D at same potential → no current through G.

Variables

P, Q, R, S = four arm resistances

Why it works

Null method measures unknown resistance accurately without galvanometer calibration.

Historical context

Balance independent of battery EMF; best when arms nearly equal.

Deep understanding

Unknown S = QR/P. Sensitive for low resistances.

2. Diagrams & Visuals

P/Q = R/S

Color-coded visual · step-by-step breakdown below

  1. Identify arms P, Q, R, S
  2. Adjust until G null
  3. P/Q = R/S
  4. Solve for unknown

3. Solved Examples

Basic

Q: P=10, Q=20, R=30 Ω.

Solution: S=60 Ω

Answer: 60 Ω

Intermediate

Q: Double Q only?

Solution: S doubles

Answer: Balance shifts

Advanced

Q: Galvanometer null?

Solution: V_B = V_D

Answer: Null method

Exam

Q: Wheatstone balance?

Solution: P/Q=R/S

Answer: Sec 17.7.1

V = (E/l) × l₁

Definition: Potentiometer principle — potential drop proportional to wire length.

Derivation

Uniform wire: potential gradient k = E/l; balance length l₁ gives V = kl₁.

Variables

E = driver cell emf · l = total wire length · l₁ = balance length

Why it works

Measures emf/potential without drawing current from source (null method).

Historical context

Driver emf must exceed unknown V; galvanometer zero at balance.

Deep understanding

k = E/l constant for uniform wire at steady current.

2. Diagrams & Visuals

l₁ V = (E/l)l₁

Color-coded visual · step-by-step breakdown below

  1. Driver cell emf E, wire length l
  2. Find balance length l₁
  3. V = (E/l)×l₁
  4. G shows zero at balance

3. Solved Examples

Basic

Q: E=2 V, l=4 m, l₁=1 m.

Solution: V=0.5 V

Answer: 0.5 V

Intermediate

Q: Double l₁?

Solution: V doubles

Answer: Linear

Advanced

Q: Why null method?

Solution: No current from unknown

Answer: Accurate emf

Exam

Q: Potentiometer V?

Solution: V=(E/l)l₁

Answer: Sec 17.8

E₁/E₂ = l₁/l₂

Definition: Comparison of two EMFs using same potentiometer wire.

Derivation

Same k = E_driver/l for both measurements; V ∝ l at balance.

Variables

l₁, l₂ = balance lengths for cells E₁, E₂

Why it works

Relative emf without knowing absolute driver calibration details.

Historical context

Same driver cell and wire required for valid comparison.

Deep understanding

Ratio method eliminates need for knowing k explicitly.

2. Diagrams & Visuals

E₁/E₂ = l₁/l₂

Color-coded visual · step-by-step breakdown below

  1. Balance cell E₁ → length l₁
  2. Balance cell E₂ → length l₂
  3. E₁/E₂ = l₁/l₂
  4. Same wire and driver

3. Solved Examples

Basic

Q: l₁=80 cm, l₂=40 cm.

Solution: E₁/E₂=2

Answer: 2:1

Intermediate

Q: E₂ twice E₁?

Solution: l₂=2l₁

Answer: Length ratio

Advanced

Q: Swap cells, same wire?

Solution: Lengths swap ratio

Answer: Consistent

Exam

Q: Compare two emfs?

Solution: E₁/E₂=l₁/l₂

Answer: Sec 17.8.3

r = (l₁/l₂ − 1) × R

Definition: Internal resistance of cell from potentiometer with shunt R.

Derivation

Open key: balance l₁ (E₁). Closed key shunting R: balance l₂ (V₁). Algebra gives r.

Variables

r = internal resistance · R = shunt resistance

Why it works

Determines r without ammeter loading errors when done as null method.

Historical context

Uses E₁/V₁ = l₁/l₂ and V₁ = E₁R/(R+r).

Deep understanding

r depends on electrode area, separation, electrolyte concentration.

2. Diagrams & Visuals

r = (l₁/l₂ − 1)R

Color-coded visual · step-by-step breakdown below

  1. Balance open circuit → l₁
  2. Close key with shunt R → l₂
  3. r = (l₁/l₂ − 1)×R
  4. Check l₁ > l₂

3. Solved Examples

Basic

Q: l₁=100, l₂=80 cm, R=10 Ω.

Solution: r=2.5 Ω

Answer: 2.5 Ω

Intermediate

Q: l₁=l₂?

Solution: r=0

Answer: Ideal cell

Advanced

Q: Why l₁ > l₂ when shunted?

Solution: V₁ < E₁

Answer: Terminal drops

Exam

Q: Internal r by potentiometer?

Solution: (l₁/l₂−1)R

Answer: Sec 17.8.4

v_d = e E τ / m

Definition: Drift velocity of conduction electrons in uniform electric field E.

Derivation

Electron accelerates between collisions; average drift from force balance with drag.

Variables

e = 1.6×10⁻¹⁹ C · τ = collision time (s) · m = electron mass

Why it works

Explains slow directed motion vs fast random thermal motion.

Historical context

Typical v_d ~ 10⁻⁴ m/s in copper; thermal ~ 10⁶ m/s.

Deep understanding

Opposite to E for electrons; combines with I = nAe v_d.

2. Diagrams & Visuals

v_d opposite E

Color-coded visual · step-by-step breakdown below

  1. Field E in conductor
  2. Collision time τ
  3. v_d = eEτ/m
  4. Use in I = nAe v_d

3. Solved Examples

Basic

Q: Double E?

Solution: v_d doubles

Answer: 2v_d

Intermediate

Q: More collisions (smaller τ)?

Solution: v_d smaller

Answer: Lower drift

Advanced

Q: No field?

Solution: v_d=0 average

Answer: Random only

Exam

Q: Drift velocity?

Solution: eEτ/m

Answer: Sec 17.9

ρ = m/(n e² τ)

Definition: Microscopic formula for electrical resistivity of metals.

Derivation

From v_d = eEτ/m and I = nAe v_d, eliminate v_d to get E/I = ρ/l/A style relation.

Variables

ρ (Ω·m) · n · τ

Why it works

Shows why more collisions (small τ) or fewer carriers raises resistivity.

Historical context

Derives macroscopic R = ρl/A from electron dynamics.

Deep understanding

Classical model; quantum corrections at low T.

2. Diagrams & Visuals

ρ = m/(ne²τ)

Color-coded visual · step-by-step breakdown below

  1. Identify n and τ
  2. ρ = m/(ne²τ)
  3. Relate to macroscopic R
  4. τ↓ → ρ↑

3. Solved Examples

Basic

Q: τ halved?

Solution: ρ doubles

Answer: 2ρ

Intermediate

Q: Higher n?

Solution: ρ lower

Answer: More carriers

Advanced

Q: Link to Ohm's law?

Solution: Combines with v_d

Answer: Microscopic

Exam

Q: Resistivity microscopic?

Solution: m/(ne²τ)

Answer: Sec 17.9

P = V I = I² R = V²/R

Definition: Electric power — rate of energy transfer or dissipation in a circuit element.

Derivation

P = dW/dt = V dq/dt = VI; substitute V=IR or I=V/R.

Variables

P (W) · 1 W = 1 J/s

Why it works

Joule heating, bulb ratings, electricity bills (kWh).

Historical context

Joule's law: heat Q = I²Rt. Domestic 1 kWh = 1 unit.

Deep understanding

Use form matching known quantities; doubling I quadruples P at fixed R.

2. Diagrams & Visuals

P = VI Heat: Q = I²Rt

Color-coded visual · step-by-step breakdown below

  1. Identify V, I, or R
  2. P = VI or I²R or V²/R
  3. Result in watts
  4. Energy E = P×t

3. Solved Examples

Basic

Q: V=12 V, I=2 A.

Solution: P=24 W

Answer: 24 W

Intermediate

Q: I=3 A, R=4 Ω.

Solution: P=36 W

Answer: 36 W

Advanced

Q: Double V, R fixed?

Solution: P quadruples

Answer: V²/R

Exam

Q: Power in resistor?

Solution: P=I²R

Answer: Sec 17.10

5. Special Features & Extras

Complete study guide for Electric Current.

Exam Tips & Tricks

  • 1 A = 1 C/s — current is charge per unit time.
  • Conventional current opposite to electron flow.
  • Resistors: series add R; parallel add 1/R (opposite to capacitors!).
  • Ohm's law: V = IR only for ohmic conductors at constant T.
  • Cell: E = V + Ir — terminal V drops when current drawn.
  • Null methods: Wheatstone bridge and potentiometer draw no balancing current.

Common Student Mistakes

  • Confusing conventional current direction with electron flow
  • Swapping series/parallel rules with capacitors
  • Using V = IR when temperature changes (filament heating)
  • Forgetting internal resistance r in cell problems
  • Wrong sign convention in Kirchhoff loop equations
  • Mixing up potentiometer driver emf with unknown cell emf

Memory Aids & Mnemonics

Resistors vs Capacitors: R series = add · C series = reciprocal
Power trio: P = VI = I²R = V²/R — pick what you know
Cell: "EMF pays the bill, Ir is the tip" → E = V + Ir
Kirchhoff: Junction = charge in/out · Loop = energy round trip = 0

Which Formula When?

  • Rate of charge flow? → I = Δq/Δt
  • Microscopic current? → I = nAe v_d
  • Resistor at fixed T? → V = IR
  • Wire dimensions? → R = ρl/A
  • Combine resistors? → series sum, parallel reciprocal
  • Real cell? → E = V + Ir
  • Complex circuit? → Kirchhoff junction + loop
  • Unknown R accurately? → Wheatstone P/Q = R/S
  • Measure emf without loading? → Potentiometer E₁/E₂ = l₁/l₂
  • Heat/power? → P = I²R

QUICK REFERENCE — Ch 17 Electric Current

I = Δq/ΔtI = n A e v_dV = R I  |  R = V/IR = ρl/AR = R₁ + R₂ + R₃ + …1/R = 1/R₁ + 1/R₂ + …ρ = ρ₀[1 + α(T − T₀)]E = V + I rΣ I_in = Σ I_outΣ (I R) = Σ EP/Q = R/SV = (E/l) × l₁E₁/E₂ = l₁/l₂r = (l₁/l₂ − 1) × Rv_d = e E τ / mρ = m/(n e² τ)P = V I = I² R = V²/R

Units: A (ampere) = C/s · Ω (ohm) = V/A · W (watt) = J/s · ρ in Ω·m

Key: I = nAe v_d · R = ρl/A · E = V+Ir · ΣIR = ΣE · P/Q = R/S

Tip: For networks, redraw series/parallel sections first; use Kirchhoff only when needed.

PYQ — Previous Year Questions

Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L17 — Electric Current only. Use Model Answer for marking points; Explanation for concept clarity.

Chapter 17 — Electric Current (L17)

13 questions · Section A (MCQ & objective) + Section B (short/long) · Sources: 312/TUS/104A, 312/MAY/204A–B, 68/ESS/1-312-A, Marking Scheme

Section A — Multiple Choice (1 mark)

PYQ1. The device used for an accurate measurement of the e.m.f. of a primary cell is — (A) galvanometer   (B) ammeter   (C) voltmeter   (D) potentiometer

1 mark · Section A Q9 · 312/MAY/204A (also Q13 · 204B)

Model Answer

Answer: (D) potentiometer

Null method — draws no current from the cell being measured.

Explanation

Voltmeter has finite resistance and draws current → terminal V < E. Potentiometer balances unknown emf against driver cell (L17 §17.8).

PYQ2. n equal resistances are first connected in series and then in parallel. The ratio of equivalent resistances Rs/Rp is — (A) 1 : n   (B) n : 1   (C) 1 : n²   (D) n² : 1

1 mark · Section A Q10 · 312/MAY/204A

Model Answer

Answer: (D) n² : 1

Rs = nR  ;  Rp = R/n → Rs/Rp = n²

Explanation

Series adds; parallel reciprocals add. For identical R, parallel gives R/n (L17 §17.3).

PYQ3. n identical cells each of emf E and internal resistance r are connected in series. The battery emf and internal resistance are respectively — (A) nE and nr   (B) nE and r/n   (C) E/n and nr   (D) E/n and r/n

1 mark · Section A Q4 · 312/MAY/204B

Model Answer

Answer: (A) nE and nr

Series: emfs add; internal resistances add.

Explanation

Same rule as resistors in series. Parallel cells would give E and r/n (L17 §17.6).

PYQ4. A resistance of 5 Ω is in the left gap and 15 Ω in the right gap of a metre-bridge. The null point from the left end is at — (A) 75 cm   (B) 60 cm   (C) 25 cm   (D) 15 cm

1 mark · Section A Q8 · 68/ESS/1-312-A

Model Answer

Answer: (C) 25 cm

R₁/R₂ = l/(100−l) → 5/15 = l/(100−l) → l = 25 cm

Explanation

Metre bridge is a Wheatstone bridge with wire arms. Balance: P/Q = R/S = l₁/l₂ (L17 §17.7.1).

PYQ5. A 100 W bulb is connected to 220 V supply. The current through the bulb is — (A) 5/11 A   (B) 10/11 A   (C) 11/5 A   (D) 11/10 A

1 mark · Section A Q14 · 68/ESS/1-312-A

Model Answer

Answer: (A) 5/11 A

I = P/V = 100/220 = 5/11 A ≈ 0.45 A

Explanation

P = VI → I = P/V. Assumes rated power at rated voltage (ohmic lamp) (L17 §17.10).

PYQ6. A 2 kg block moves at constant velocity 5 m·s⁻¹ under a constant 3 N force. Power dissipated against friction is — (A) zero   (B) 15 W   (C) −15 W   (D) 30 W

1 mark · Section A Q13 · 68/ESS/1-312-A

Model Answer

Answer: (B) 15 W

Constant v ⇒ friction = 3 N opposite motion. P = Fv = 3 × 5 = 15 W

Explanation

Net force zero at constant velocity. Applied 3 N equals friction. Electrical analogue: P = I²R heat dissipation (L17 §17.10).

PYQ7. To obtain the maximum resistance, three resistors r₁, r₂ and r₃ should be connected as — (diagram options A–D showing series/parallel combinations)

1 mark · Section A Q9 · 312/TUS/104A

Model Answer

All three in series (option with series connection).

Rmax = r₁ + r₂ + r₃

Explanation

Maximum resistance when no parallel shunting — full series combination.

PYQ8. A wire of length L and diameter D will have minimum resistance when its length and diameter are — (A) L and D   (B) L and D/2   (C) 2L and 2D   (D) L/2 and 2D

1 mark · Section A Q9 (OR) · 312/TUS/104A

Model Answer

Answer: (A) L and D (shortest length, largest area → minimum R = ρl/A)

Explanation

R ∝ l and R ∝ 1/A ∝ 1/D². Minimum R for given material: smallest l, largest D.

PYQ9. Fill in the blanks (any two): (a) The other name for joule per second is _____   (b) 1 kWh is the unit of _____   (c) 1 horsepower = _____ watt (approx.)

2 marks (1×2) · Section A Q20 · 312/TUS/104A

Model Answer

(a) watt (power)

(b) energy (electrical energy consumed)

(c) 746 W (accept 750 W)

Explanation

P = W/t → 1 W = 1 J/s. Domestic bill uses kWh; 1 kWh = 3.6×10⁶ J (Section B Q33).

PYQ10. Match circuit combination with property (any two): (a) Series resistors → ?   (b) Parallel resistors → ?   (c) Balanced Wheatstone bridge → ?   Options: (i) same current   (ii) same potential   (iii) P/Q = R/S   (iv) galvanometer shows deflection

2 marks (1×2) · Section A Q21 · 312/MAY/204A

Model Answer

(a) Series ↔ (i) same current

(b) Parallel ↔ (ii) same potential

(c) Balanced bridge ↔ (iii) P/Q = R/S (null galvanometer)

Explanation

Unbalanced bridge → galvanometer deflects. Series adds R; parallel adds 1/R (L17 §17.3, §17.7.1).

PYQ11. Show that 1 kWh = 3.6 × 10⁶ J.

2 marks · Section B Q33 · Marking Scheme

Model Answer

1 kWh = 1 kW × 1 h = 1000 W × 3600 s

= 1000 J/s × 3600 s = 3.6 × 10⁶ J

Explanation

1 unit on electricity bill = 1 kWh. Links power (kW) and energy (J) — board marking scheme derivation.

PYQ12. Match the circuit in Column—I with its equivalent resistance in Column—II (any two): Options include 6.0 Ω, 0.5 Ω, 1.5 Ω, 2.2 Ω for series/parallel resistor diagrams.

2 marks (1×2) · Section A Q28 · 312/TUS/104A

Model Answer

Match each circuit diagram to calculated Req using series R = R₁+R₂ and parallel 1/R = 1/R₁+1/R₂.

Board paper gives four specific values — compute for each diagram shown in paper.

Explanation

TUS 104A Q28 circuit-resistance match. Refer original paper diagrams for exact pairings.

PYQ13. A galvanometer of coil resistance 12 Ω gives full-scale deflection for 2.5 mA. How will you convert it into (a) an ammeter of range 0–2 A and (b) a voltmeter of range 0–10 V?

5 marks · Section B Q43 OR · 312/MAY/204A

Model Answer

(a) Ammeter: Connect low resistance S in parallel (shunt).

IgRg = (I − Ig)S → S = IgRg/(I − Ig) = 0.0025×12/(2−0.0025) ≈ 0.015 Ω

(b) Voltmeter: Connect high resistance R in series.

V = Ig(Rg + R) → R = V/Ig − Rg = 10/0.0025 − 12 = 3988 Ω

Explanation

Shunt bypasses most current; series multiplier drops most voltage. Ammeter has very low resistance; voltmeter very high resistance.

Problem Solving — L17 Electric Current

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.

Question 1 of 6Current

5.0 C crosses a section in 2.0 s. Find average current.

I = q/t
I = nAve

Solution — step by step with formulas

  1. I = 2.5 A.

Final answer: I = 2.5 A

Formulas used in this problem

I = q/t
I = nAve

Textbook formal language

Current is rate of flow of charge through a cross-section.

Working formula set for this problem: I = q/t; I = nAve. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Charge per second: 5/2 = 2.5 amperes.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Current definition

Conventional current is direction of positive flow.

Link to chapter notes (L17 — Current definition): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: I = q/t; I = nAve. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write I = q/t; I = nAve before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 2 of 6Ohm

A 12 V battery drives 0.50 A through a resistor. Find R and power dissipated.

V = IR

Solution — step by step with formulas

  1. R = V/I = 24 Ω.
  2. P = VI = 6.0 W (or I²R).

Final answer: R = 24 Ω; P = 6 W

Formulas used in this problem

V = IR

Textbook formal language

Ohm’s law: V proportional to I for ohmic conductors at fixed T.

Working formula set for this problem: V = IR. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Resistance is volts per amp; power is heat per second in the resistor.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Ohm’s law

Non-ohmic devices (diodes) do not give linear V–I.

Link to chapter notes (L17 — Ohm’s law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: V = IR. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write V = IR before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 3 of 6Resistivity

Wire: ρ = 1.7×10⁻⁸ Ω·m, L = 2.0 m, A = 1.0×10⁻⁶ m². Find R.

R = ρL/A

Solution — step by step with formulas

  1. R = ρL/A = 0.034 Ω.

Final answer: R = 0.034 Ω

Formulas used in this problem

R = ρL/A

Textbook formal language

Resistance depends on material (ρ), length, and cross-section.

Working formula set for this problem: R = ρL/A. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Long thin wires resist more; copper has small ρ.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Resistivity

ρ often rises with temperature for metals.

Link to chapter notes (L17 — Resistivity): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: R = ρL/A. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write R = ρL/A before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 4 of 6Series/parallel R

Two 6 Ω resistors: R_eq series and parallel?

R_s = ΣR
1/R_p = Σ1/R

Solution — step by step with formulas

  1. Series 12 Ω; parallel 3 Ω.

Final answer: 12 Ω series; 3 Ω parallel

Formulas used in this problem

R_s = ΣR
1/R_p = Σ1/R

Textbook formal language

Series: same current; parallel: same voltage.

Working formula set for this problem: R_s = ΣR; 1/R_p = Σ1/R. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Chain adds resistance; side-by-side cuts resistance.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Resistor networks

Use Kirchhoff for complex circuits.

Link to chapter notes (L17 — Resistor networks): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: R_s = ΣR; 1/R_p = Σ1/R. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write R_s = ΣR; 1/R_p = Σ1/R before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 5 of 6Kirchhoff

State Kirchhoff’s junction and loop rules briefly.

ΣI = 0 (junction)
Σε = ΣIR (loop)

Solution — step by step with formulas

  1. Junction: charge conservation, currents balance.
  2. Loop: energy conservation, sum of PD = 0.

Final answer: ΣI_in = ΣI_out; ΣΔV around loop = 0

Formulas used in this problem

ΣI = 0 (junction)
Σε = ΣIR (loop)

Textbook formal language

Kirchhoff rules implement conservation laws in steady circuits.

Working formula set for this problem: ΣI = 0 (junction); Σε = ΣIR (loop). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

What flows in must flow out; walking a loop, gains and drops cancel.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Kirchhoff’s laws

Essential for multi-loop circuits with several batteries.

Link to chapter notes (L17 — Kirchhoff’s laws): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: ΣI = 0 (junction); Σε = ΣIR (loop). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write ΣI = 0 (junction); Σε = ΣIR (loop) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 6 of 6Emf/internal r

Cell ε = 2.0 V, r = 0.50 Ω, external R = 1.5 Ω. Find current and terminal voltage.

V = ε − Ir

Solution — step by step with formulas

  1. I = ε/(R+r) = 1.0 A.
  2. V = IR = 1.5 V (or ε−Ir).

Final answer: I = 1.0 A; V = 1.5 V

Formulas used in this problem

V = ε − Ir

Textbook formal language

Terminal voltage is less than emf by the internal drop Ir when current is drawn.

Working formula set for this problem: V = ε − Ir. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Battery’s own resistance eats 0.5 V; you measure 1.5 V at terminals.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Cell with internal resistance

On open circuit I = 0 and V = ε.

Link to chapter notes (L17 — Cell with internal resistance): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: V = ε − Ir. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write V = ε − Ir before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).