Current Electricity - Class 12 Physics

Class 12 Physics | Unit I — Current Electricity

Chapter 3: Current Electricity

Ohm's Law • Drift Velocity • Resistivity • Kirchhoff's Laws • Wheatstone Bridge • Potentiometer

1. Electric Current and Drift Velocity

Electric current (I): Rate of flow of charge. I = Q/t (A = C/s). Conventional current flows from + to − (outside battery); electron flow is opposite direction. Current is a scalar (but has direction → treated as vector at junction).

I = nqAv_d
n = number density of free electrons (m⁻³), q = charge of electron = 1.6×10⁻¹⁹ C, A = cross-sectional area (m²), v_d = drift velocity (m/s)

Drift velocity: v_d = eEτ/m = eVτ/(mL)
τ = relaxation time (average time between collisions, ~10⁻¹⁴ s), E = electric field, m = mass of electron

Drift velocity is very small (~10⁻⁴ m/s to 10⁻³ m/s) compared to random motion (~10⁶ m/s).
Current flows at nearly speed of light (electric field propagates at c) even though drift velocity is slow — like water in a full pipe.
When potential difference applied: free electrons drift slowly opposite to E field, giving net current in direction of E.
v_d ∝ E ∝ V (for same L): doubling voltage doubles drift velocity.
Current density: J = I/A = nqv_d (A/m²). Vector quantity → J = σE.

⚠️ NEET: v_d = eEτ/m. I = nqAv_d. If A doubles and I same → v_d halves. If L doubles and V same → E halves → v_d halves. Drift velocity is in direction opposite to E (electrons are negative). Relaxation time τ ≈ 10⁻¹⁴ s.

2. Ohm's Law and Resistance

Ohm's Law: For a conductor at constant temperature, the current (I) through it is directly proportional to the potential difference (V) across its ends. V = IR. R = resistance (Ω).

R = ρL/A
ρ = resistivity (Ω·m), L = length, A = cross-sectional area

Temperature dependence: ρ = ρ₀[1 + α(T − T₀)]
α = temperature coefficient of resistivity (°C⁻¹). For metals: α > 0 (ρ increases with T). For semiconductors/insulators: α < 0 (ρ decreases with T).

2.1 Resistivity of Materials

Material Type	Resistivity (Ω·m)	α	Examples
Conductors (metals)	10⁻⁸ to 10⁻⁶	> 0	Copper (1.7×10⁻⁸), Silver, Aluminium
Semiconductors	10⁻⁵ to 10⁶	< 0	Si, Ge
Insulators	10⁸ to 10¹⁷	< 0	Glass, Rubber, Teflon
Alloys (special)	Moderate	≈ 0	Manganin, Nichrome, Constantan

Nichrome: High resistivity, high melting point → used in heaters, toasters.
Manganin/Constantan: Very low α → used in standard resistors (accuracy).
Superconductors: ρ = 0 below critical temperature T_c. No resistance, no energy loss. Example: Mercury below 4.2 K.

2.2 Ohmic vs Non-Ohmic Conductors

Ohmic (linear)	Non-Ohmic (non-linear)
V-I graph = straight line through origin	V-I graph is curved (not linear)
R is constant (independent of V, I)	R changes with V or I or direction
Examples: metallic resistors, wires	Examples: diode, LED, transistor, electrolyte, thyristor

2.3 Resistors in Series and Parallel

Series: R_eff = R₁ + R₂ + R₃ + ... (current same, voltage divides)
Parallel: 1/R_eff = 1/R₁ + 1/R₂ + 1/R₃ + ... (voltage same, current divides)
For 2 resistors in parallel: R_eff = R₁R₂/(R₁+R₂) (product over sum)

Series combination: R_eff > largest individual R.
Parallel combination: R_eff < smallest individual R.
Power: P = VI = I²R = V²/R. For series: P ∝ R (more power across larger R). For parallel: P ∝ 1/R (more power through smaller R).

3. EMF, Terminal Voltage and Internal Resistance

EMF (ε): The work done per unit charge by the cell to move charge from negative to positive terminal through the battery (against electric force). Unit: Volt (V). EMF = open-circuit voltage.

Terminal voltage under load: V = ε − Ir (during discharge / when current is drawn)
During charging: V = ε + Ir (terminal voltage exceeds EMF)
r = internal resistance of cell (Ω)

Current in circuit: I = ε / (R + r)
Short circuit current (R = 0): I_sc = ε/r

Terminal voltage < EMF when discharging (energy lost in internal resistance).
When cell is on open circuit (I = 0): V = ε (terminal voltage = EMF exactly).
Power delivered to external R: P_ext = I²R = ε²R/(R+r)².
Maximum power transfer theorem: P_max transferred to external R when R = r (internal = external resistance). P_max = ε²/(4r).

3.1 Cells in Series and Parallel

Combination	Net EMF	Net Internal Resistance	Best for
Series (n cells)	nε	nr	High resistance external circuits (R >> r)
Parallel (n cells)	ε	r/n	Low resistance external circuits (R << r)
Mixed (m rows × n columns)	nε	nr/m	Balanced when R = nr/m

⚠️ NEET (2014, 2017, 2019, 2021): V = ε − Ir (discharge). V = ε + Ir (charge). I = ε/(R+r). Open circuit: V = ε. Short circuit: I = ε/r. Max power to R when R = r. Series cells: EMF adds, r adds. Parallel cells: EMF same, r divides.

4. Kirchhoff's Laws

4.1 KCL — Kirchhoff's Current Law (Junction Rule)

KCL: The algebraic sum of all currents at any junction (node) in a circuit is zero. Equivalently: ΣI_in = ΣI_out. Based on conservation of electric charge.

At a junction: current entering = current leaving. No charge accumulates at a junction.

4.2 KVL — Kirchhoff's Voltage Law (Loop Rule)

KVL: The algebraic sum of all potential differences (EMFs and voltage drops) around any closed loop in a circuit is zero. Based on conservation of energy.

Σε − ΣIR = 0 (around any closed loop)
Sign convention: EMF is +ve if we go from − to + terminal through the source; voltage drop IR is −ve in direction of current.

KCL: Based on charge conservation. At a node, ∑I = 0.
KVL: Based on energy conservation. In any closed loop, ∑V = 0.
Used solves complex networks (multiple loops, multiple batteries) where series/parallel rules don't apply directly.

5. Wheatstone Bridge

Wheatstone Bridge: A circuit arrangement of four resistors (P, Q, R, S) with a battery and a galvanometer (G), used to measure unknown resistance accurately. Devised by Wheatstone (1843), principle discovered by Christie (1833).

Balanced condition (G shows zero deflection, I_g = 0):
P/Q = R/S or equivalently PS = QR
If three resistors known, the fourth (unknown) can be calculated.

In balanced state: no current through galvanometer branch → sensitive galvanometer is protected.
Galvanometer deflects when bridge is unbalanced: P/Q ≠ R/S.
Metre bridge (Slide Wire Bridge): practical application of Wheatstone bridge. Uses uniform resistance wire of 100 cm. R_unknown/R_known = l/(100−l) where l = balancing length.
Sensitivity: increases when all four arms have similar resistance values.
Application: measurement of resistance, strain gauge, temperature sensors (thermistor bridges).

5.1 Metre Bridge Formula

R_X = R_S × l / (100 − l)
l = balancing length (cm), R_S = known resistance in resistance box
Also used to find specific resistance: ρ = RA/L

6. Potentiometer

Potentiometer: An instrument for measuring potential difference accurately without drawing any current from the circuit being measured. Works on the principle that potential drop is directly proportional to length for a uniform wire carrying steady current: V ∝ l.

Potential gradient: k = V/L (V/m), where V = driving voltage across full wire of length L.
At balance: ε = kl (no current through galvanometer; null deflection)
EMF comparison: ε₁/ε₂ = l₁/l₂
Internal resistance: r = R(l₁ − l₂)/l₂ (l₁=open circuit length, l₂=loaded circuit length)

6.1 Applications of Potentiometer

Comparison of EMFs: ε₁/ε₂ = l₁/l₂ (balance lengths for each cell).
Measurement of internal resistance: r = (l₁−l₂)R/l₂.
Advantages over voltmeter: Does not draw any current from the source → no loading effect → more accurate. Acts as a null-deflection instrument.
Sensitivity increased by: using longer wire (smaller k), increasing total resistance, using more accurate galvanometer.

6.2 Potentiometer vs Voltmeter

Feature	Potentiometer	Voltmeter
Current drawn	Zero (at balance)	Small current drawn
Accuracy	Very high (null method)	Lower (loading effect)
Nature	Null-deflection instrument	Deflection instrument
Measures	EMF and potential difference	Terminal voltage only
Range	Limited by wire EMF	Can be extended

⚠️ NEET (2013, 2016, 2018, 2020, 2022): Wheatstone balance: P/Q = R/S. Metre bridge: R_x = R_s×l/(100−l). Potentiometer: ε₁/ε₂ = l₁/l₂. Internal resistance: r = (l₁−l₂)R/l₂. Potentiometer draws NO current at balance (more accurate than voltmeter). Sensitivity ↑ → longer wire / lower k.

7. Electrical Energy and Power

Power: P = VI = I²R = V²/R
Electrical energy: W = Pt = VIt = I²Rt = V²t/R
Unit of energy: Joule (J) or kWh. 1 kWh = 3.6×10⁶ J = 3.6 MJ.

Heat generated (Joule's law): H = I²Rt (in joules). This is the Joule heating effect.
Applications of heating effect: electric bulb (tungsten filament), electric iron, electric heater (nichrome wire), fuse.
Fuse wire: Made of low melting point material. Melts when current exceeds rated value → breaks circuit → protects appliances. Made of Pb-Sn alloy.
In series combination: same I → power P = I²R → max power in largest R.
In parallel combination: same V → power P = V²/R → max power in smallest R.
Efficiency: η = P_external/P_total = R/(R+r) × 100%.

🎓 NEET Previous Year Questions

Q1. [NEET 2022] If the drift velocity of electrons in a conductor is v_d when connected to a battery of EMF ε, what happens to v_d if the length of the conductor is doubled while V remains same?

Answer v_d = eVτ/(mL). If L doubles and V same: E = V/L halves → v_d halves. v_d becomes v_d/2. (Also: R doubles → I halves → I = nqAv_d → v_d halves.)

Q2. [NEET 2021] A cell of EMF 1.5 V and internal resistance 1 Ω is connected to a resistor of 4 Ω. Find the terminal voltage.

Answer I = ε/(R+r) = 1.5/(4+1) = 0.3 A. Terminal voltage V = ε − Ir = 1.5 − 0.3×1 = 1.2 V.

Q3. [NEET 2020] In a Wheatstone bridge, P = 10 Ω, Q = 15 Ω, S = 30 Ω and R is unknown. For balance:

Answer At balance: P/Q = R/S ⇒ R = PS/Q = 10×30/15 = R = 20 Ω.

Q4. [NEET 2019] Two cells of EMF ε and 2ε and internal resistances r and 2r are connected in series. Total EMF and internal resistance:

Answer In series: EMF_total = ε + 2ε = 3ε. r_total = r + 2r = 3r. (Assuming both cells assist: + + −). Note: if cells oppose each other, net EMF = 2ε − ε = ε.

Q5. [NEET 2018] In a potentiometer, the balancing length for cell of EMF 1.2 V is 48 cm and for cell of unknown EMF is 36 cm. Find the unknown EMF.

Answer ε₂/ε₁ = l₂/l₁ ⇒ ε₂ = ε₁ × l₂/l₁ = 1.2 × 36/48 = 1.2 × 0.75 = 0.9 V.

Q6. [NEET 2017] Three resistors 1 Ω, 2 Ω, 3 Ω are connected in parallel. Which carries maximum current when connected to 6 V battery?

Answer Voltage across each = 6 V. I = V/R. I₁ = 6/1 = 6A, I₂ = 6/2 = 3A, I₃ = 6/3 = 2A. 1 Ω resistor carries maximum current (6A). In parallel, smaller R → more current.

💡 Rapid Revision — Key Formulas

I = nqAv_d | v_d = eEτ/m | J = I/A = σE
V = IR | R = ρL/A | ρ = ρ₀[1+α(T−T₀)]
V = ε − Ir (discharge) | I = ε/(R+r) | Max power when R = r
KCL: ∑I = 0 at junction | KVL: ∑V = 0 around loop
Wheatstone: P/Q = R/S | Metre bridge: R_x = R_sl/(100−l)
Potentiometer: ε₁/ε₂ = l₁/l₂ | r = (l₁−l₂)R/l₂
P = VI = I²R = V²/R | H = I²Rt | 1 kWh = 3.6×10⁶ J
Series: same I, V divides | Parallel: same V, I divides

Numericals - Current Electricity - Class 12

CLASS 12 PHYSICS | NCERT SOLUTIONS

Chapter 3 — Current Electricity

25 NCERT Exercise & Exemplar Questions — Step-by-Step Solutions

📝 NCERT Exercise Questions (3.1 – 3.16)

3 MarksQ1 (Ex 3.1). The storage battery of a car has an EMF of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

✓ Solution
Given: ε = 12 V, r = 0.4 Ω
Maximum current = short-circuit current (R_ext = 0):

I_max = ε/r = 12/0.4 = 30 A

This is the current when the battery terminals are short-circuited. In practice such high current would damage the battery quickly.

3 MarksQ2 (Ex 3.2). A battery of EMF 10 V and internal resistance 3 Ω is connected to a resistor. The current in the circuit is 0.5 A. Find (a) resistance of resistor, (b) terminal voltage of battery.

✓ Solution
Given: ε = 10 V, r = 3 Ω, I = 0.5 A

(a) Resistance R:

ε = I(R + r) ⇒ R = ε/I − r = 10/0.5 − 3 = 20 − 3 = 17 Ω

(b) Terminal voltage:

V = ε − Ir = 10 − 0.5×3 = 10 − 1.5 = 8.5 V

Or: V = IR = 0.5 × 17 = 8.5 V ✓

3 MarksQ3 (Ex 3.3). Three resistors 1 Ω, 2 Ω, 3 Ω are combined in (a) series and (b) parallel. What is the total resistance in each case?

✓ Solution
(a) Series:

R_eff = R₁ + R₂ + R₃ = 1 + 2 + 3 = 6 Ω

(b) Parallel:

1/R_eff = 1/1 + 1/2 + 1/3 = 6/6 + 3/6 + 2/6 = 11/6

R_eff = 6/11 = 0.545 Ω ≈ 6/11 Ω

3 MarksQ4 (Ex 3.4). At room temperature (27°C) the resistivity of a wire is 1.7×10⁻⁸ Ω·m. At what temperature would this be doubled? (α = 6.0×10⁻³ /°C for copper)

✓ Solution
Given: ρ₀ = 1.7×10⁻⁸ Ω·m at T₀=27°C, ρ = 2×1.7×10⁻⁸ Ω·m, α = 6.0×10⁻³ /°C

ρ = ρ₀[1 + α(T−T₀)] ⇒ 2 = 1 + 6×10⁻³(T−27)

1 = 6×10⁻³(T−27) ⇒ T−27 = 1/(6×10⁻³) ≈ 166.7°C
T = 27 + 166.7 ≈ 193.7°C ≈ 194°C

3 MarksQ5 (Ex 3.5). A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6×10⁻⁷ m² and its resistance is measured to be 5 Ω. What is the resistivity of the wire?

✓ Solution
Given: L = 15 m, A = 6×10⁻⁷ m², R = 5 Ω

ρ = RA/L = (5 × 6×10⁻⁷) / 15 = 30×10⁻⁷ / 15 = 2×10⁻⁷ Ω·m

This value corresponds to an alloy (between conductor and semiconductor range).

3 MarksQ6 (Ex 3.6). A silver wire has a resistance of 2.1 Ω at 27.5°C and a resistance of 2.7 Ω at 100°C. Find the temperature coefficient of resistance α.

✓ Solution
Given: R₁ = 2.1 Ω at T₁ = 27.5°C, R₂ = 2.7 Ω at T₂ = 100°C
R₂ = R₁[1 + α(T₂−T₁)]

2.7 = 2.1[1 + α(100−27.5)] ⇒ α = (2.7/2.1 − 1) / 72.5 = (0.2857) / 72.5

α = 3.94×10⁻³ /°C ≈ 3.9×10⁻³ °C⁻¹

3 MarksQ7 (Ex 3.7). A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. Find: (a) steady-state temperature of element (b) steady-state resistance. [ρ₂₀ = 1.00×10⁻⁶ Ω·m, α_Nichrome = 1.70×10⁻⁴/°C]

✓ Solution
R at 20°C (initial): R₂₀ = 230/3.2 = 71.87 Ω
Steady-state resistance: R_T = 230/2.8 = 82.14 Ω

R_T = R₂₀[1 + α(T−20)] ⇒ 82.14 = 71.87[1 + 1.7×10⁻⁴(T−20)]

1 + 1.7×10⁻⁴(T−20) = 82.14/71.87 = 1.1429
T−20 = 0.1429/(1.7×10⁻⁴) = 840.6°C
T ≈ 861°C (steady state temperature of nichrome element)

3 MarksQ8 (Ex 3.8). Determine the current in each branch of the Wheatstone bridge-like network: AB = 1Ω, AD = 2Ω, BC = 2Ω, DC = 4Ω, AC (battery branch) = 6V, BD (galvanometer) = 0 (balanced). Find currents.

✓ Solution
Bridge is balanced when P/Q = R/S: Check: AB/BC = 1/2 and AD/DC = 2/4 = 1/2 → Bridge IS balanced. No current through galvanometer (BD).
Path ABС: R = 1 + 2 = 3 Ω; I₁ = 6/3 = 2 A (through AB and BC)
Path ADC: R = 2 + 4 = 6 Ω; I₂ = 6/6 = 1 A (through AD and DC)
Total current from battery = 2 + 1 = 3 A. Current through galvanometer = 0.

3 MarksQ9 (Ex 3.9). A metre bridge has its balance point at 39.5 cm when the resistance in the gap Q is 12.5 Ω. What is the unknown resistance R?

✓ Solution
Given: l = 39.5 cm, Q = 12.5 Ω

R/Q = l/(100−l) ⇒ R = Q × l/(100−l) = 12.5 × 39.5/(100−39.5)

R = 12.5 × 39.5/60.5 = 12.5 × 0.6529 = 8.16 Ω

3 MarksQ10 (Ex 3.10). In a potentiometer arrangement, a cell of EMF 1.25 V gives a balance at a length of 31.5 cm. Find (a) potential drop per cm in the wire, (b) EMF of another cell that gives balance point at 64.0 cm.

✓ Solution
(a) Potential gradient:

k = ε₁/l₁ = 1.25/31.5 = 0.0397 V/cm ≈ 0.04 V/cm

(b) Unknown EMF:

ε₂ = k × l₂ = 0.0397 × 64.0 = 2.54 V

OR: ε₂/ε₁ = l₂/l₁ ⇒ ε₂ = 1.25 × 64/31.5 = 2.54 V ✓

3 MarksQ11 (Ex 3.11). Find the current drawn from the battery by the network: 5 Ω and 10 Ω in parallel, in series with 3 Ω, connected to a battery of 12V, r = 1 Ω.

✓ Solution
Parallel combination: R_p = (5×10)/(5+10) = 50/15 = 10/3 Ω
Total external resistance: R = R_p + 3 = 10/3 + 3 = 10/3 + 9/3 = 19/3 Ω
Total circuit R = R_ext + r = 19/3 + 1 = 19/3 + 3/3 = 22/3 Ω

I = ε/(R+r) = 12 / (22/3) = 12 × 3/22 = 36/22 = 1.636 A ≈ 1.64 A

3 MarksQ12 (Ex 3.12). A potentiometer wire is 100 cm long and has a total PD of 6 V across it. Find (a) PD per cm, (b) length for which cell of EMF 1.5 V would balance.

✓ Solution
(a) Potential gradient:

k = V/L = 6/100 = 0.06 V/cm

(b) Balance length for ε = 1.5 V:

l = ε/k = 1.5/0.06 = 25 cm

3 MarksQ13 (Ex 3.13). The number density of free electrons in a copper conductor is 8.5×10²⁸/m³. How long does an electron take to drift from one end to the other of a wire of length 3 m, carrying 3.0 A of current? (Cross-section area = 2×10⁻⁶ m²)

✓ Solution
Given: n = 8.5×10²⁸ m⁻³, L = 3 m, I = 3 A, A = 2×10⁻⁶ m², e = 1.6×10⁻¹⁹ C

v_d = I/(nqA) = 3 / (8.5×10²⁸ × 1.6×10⁻¹⁹ × 2×10⁻⁶)

v_d = 3 / (2.72×10⁴) = 1.103×10⁻⁴ m/s

t = L/v_d = 3 / (1.103×10⁻⁴) = 2.72×10⁴ s ≈ 7.56 hours

This shows how slow drift velocity really is (electrons take hours to drift 3 m!).

3 MarksQ14 (Ex 3.14). The earth's surface has a surface charge density of −10⁻⁹ C/m². The current of 1800 A enters the earth at the North Pole per second. Calculate the time required to neutralise the charge. (Radius of earth = 6400 km)

✓ Solution
Surface charge density σ = −10⁻⁹ C/m²; R_e = 6.4×10⁶ m
Total charge: Q = σ × 4πR_e² = 10⁻⁹ × 4π × (6.4×10⁶)²
Q = 10⁻⁹ × 4π × 4.096×10¹³ = 10⁻⁹ × 5.15×10¹⁴ = 5.15×10⁵ C

t = Q/I = 5.15×10⁵ / 1800 = 286 s ≈ 4.8 minutes

3 MarksQ15 (Ex 3.15). For a cell, the terminal voltage is 1.8 V when current is 0.5 A, and terminal voltage is 2.0 V when current is 0.1 A. Find the EMF and internal resistance of the cell.

✓ Solution
V₁ = ε − I₁r ⇒ 1.8 = ε − 0.5r ... (1)
V₂ = ε − I₂r ⇒ 2.0 = ε − 0.1r ... (2)
(2) − (1): 0.2 = 0.4r ⇒ r = 0.5 Ω
From (2): ε = 2.0 + 0.1×0.5 = 2.0 + 0.05 = ε = 2.05 V

3 MarksQ16 (Ex 3.16). Two wires A and B of the same material with lengths in ratio 1:2 and radii in ratio 1:2 are connected in series to a battery. The ratio of potential differences across A and B is:

✓ Solution
R = ρL/A = ρL/(πr²)
R_A/R_B = (L_A/L_B) × (r_B/r_A)² = (1/2) × (2/1)² = (1/2) × 4 = 2:1
In series, same current flows. V = IR. So V_A/V_B = R_A/R_B = 2:1.

🌟 Additional / Exemplar Questions (Q17 – Q25)

3 MarksQ17. Six resistors of 3 Ω each are connected in the form of a Wheatstone bridge. Find the effective resistance between two opposite corners (A and C, with battery across AC and galvanometer across BD).

✓ Solution
The bridge is balanced (all R = 3Ω, so P/Q = R/S = 1, all balanced). No current through BD branch.
Path A→B→C: 3+3 = 6Ω. Path A→D→C: 3+3 = 6Ω. These two paths in parallel:
R_eff = 6×6/(6+6) = 36/12 = 3 Ω.

3 MarksQ18. 20 cells each of EMF 2 V and internal resistance 0.5 Ω are to be connected to supply current to an external resistance R = 10 Ω. Should they be connected in series or parallel? Find current in each case.

✓ Solution
Series (n = 20): EMF = 20×2 = 40V; r_total = 20×0.5 = 10Ω
I_series = 40/(10+10) = 40/20 = 2.0 A

Parallel (n = 20): EMF = 2V; r_eff = 0.5/20 = 0.025Ω
I_parallel = 2/(10+0.025) ≈ 2/10.025 ≈ 0.2 A

Series gives more current (2 A) when R > r/n. Use series when external R is large.

3 MarksQ19. A potentiometer wire of length 1 m is connected to a driver battery of 2 V. When a cell of EMF ε gives balance at 60 cm and when a shunt resistance R = 5 Ω is connected across cell, balance shifts to 40 cm. Find: EMF and internal resistance of cell.

✓ Solution
k = 2/100 cm = 0.02 V/cm (potential gradient)
EMF ε = k × 60 = 0.02 × 60 = 1.2 V
With shunt: terminal voltage = k × 40 = 0.02 × 40 = 0.8 V
Current through shunt: I = V/R = 0.8/5 = 0.16 A
V = ε − Ir ⇒ 0.8 = 1.2 − 0.16r ⇒ r = 0.4/0.16 = 2.5 Ω

3 MarksQ20. An electric bulb is rated 220 V, 100 W. (a) What is its resistance? (b) What current flows through it at rated voltage? (c) How much heat is generated in 10 minutes?

✓ Solution
(a) Resistance:

R = V²/P = (220)²/100 = 48400/100 = 484 Ω

(b) Current:

I = P/V = 100/220 = 0.455 A

Or I = V/R = 220/484 = 0.455 A ✓

(c) Heat in 10 min (600 s):

H = Pt = 100 × 600 = 60,000 J = 60 kJ

3 MarksQ21. A copper wire of length 2 m and cross-section 1 mm² has a resistance of 34 mΩ. Find the resistivity of copper.

✓ Solution
Given: L = 2 m, A = 1 mm² = 1×10⁻⁶ m², R = 34×10⁻³ Ω

ρ = RA/L = (34×10⁻³ × 10⁻⁶) / 2 = (34×10⁻⁹) / 2

ρ = 1.7×10⁻⁸ Ω·m (consistent with standard value for copper)

3 MarksQ22. Two batteries of EMF 6 V (r₁=1Ω) and 2 V (r₂=1Ω) are connected in series with external R = 10Ω. Find current and terminal voltage of each cell.

✓ Solution
Cells aid each other: net EMF = 6+2 = 8 V; r_total = 1+1 = 2Ω

I = ε_net/(R+r_total) = 8/(10+2) = 8/12 = 2/3 A ≈ 0.667 A

Terminal V of 6V cell: V₁ = ε₁ − Ir₁ = 6 − 0.667×1 = 5.33 V
Terminal V of 2V cell: V₂ = ε₂ − Ir₂ = 2 − 0.667×1 = 1.33 V

3 MarksQ23. Three identical bulbs each rated 60 W, 250 V are connected: (a) all three in series across 250 V, (b) all three in parallel across 250 V. Find total power consumed in each case.

✓ Solution
R of each bulb = V²/P = (250)²/60 = 1041.7 Ω

(a) Series (same current):
Total R = 3×1041.7 = 3125 Ω; I = 250/3125 = 0.08 A
P_total = I²×3R = (0.08)²×3125 = 20 W (or P/3 = 60/3 = 20 W)

(b) Parallel (same voltage):
P_total = 3×60 = 180 W (each gets full 250V; full rated power)

3 MarksQ24. An electron in a copper wire moves with drift velocity 1 mm/s. If n = 8.5×10²⁸/m³ and area = 1 mm², find the current carried by the wire.

✓ Solution
Given: v_d = 1 mm/s = 10⁻³ m/s, n = 8.5×10²⁸ m⁻³, A = 10⁻⁶ m²

I = nqAv_d = 8.5×10²⁸ × 1.6×10⁻¹⁹ × 10⁻⁶ × 10⁻³

I = 8.5 × 1.6 × 10^{28−19−6−3} = 13.6 × 10⁰ = 13.6 A

3 MarksQ25. Derive the condition of balance for a Wheatstone bridge and explain why a potentiometer gives a more accurate measurement of EMF than a voltmeter.

✓ Solution
Wheatstone Bridge Balance:
Four resistors P, Q, R, S in a diamond/bridge arrangement. Battery across AC; galvanometer across BD. At balance: I_g = 0 (no current through galvanometer).
Applying KCL: I₁ flows through P and R; I₂ flows through Q and S (since I_g=0).
Applying KVL to loop ABD: I₁P = I₂Q ... (i)
Applying KVL to loop BCD: I₁R = I₂S ... (ii)
Dividing (i)/(ii): P/Q = R/S — the balance condition.

Why potentiometer > voltmeter for EMF:
A voltmeter has finite resistance → draws current from cell → causes voltage drop across internal resistance → reads terminal voltage (not EMF). A potentiometer at balance draws zero current from the cell being measured → no voltage drop across r → reads true EMF. The null-method is inherently more accurate.

✍ Score Guide — 25 Questions
NCERT Ex 3.1–3.16: 3 marks each — 48 marks | Additional Q17–Q25: 3 marks each — 27 marks | Grand Total: 75 marks

Formula Capsule - Current Electricity - Class 12

CLASS 12 PHYSICS | FORMULA CAPSULE

Current Electricity

Chapter 3 — All Key Formulas & High-Yield Facts for NEET/JEE

📅 Section A — Current, Drift Velocity & Resistance

Quantity	Formula	Key Note
Electric current	I = Q/t = nqAv_d	n=electron density, A=area, v_d=drift velocity
Current density	J = I/A = nqv_d = σE	σ = conductivity = 1/ρ
Drift velocity	v_d = eEτ/m = eVτ/(mL)	τ = relaxation time (~10⁻¹⁴ s)
Ohm's Law	V = IR	Valid for ohmic conductors only
Resistance	R = ρL/A	Depends on material, length, cross-section
Resistivity (temp)	ρ = ρ₀[1 + α(T−T₀)]	Metals: α>0; Semiconductors: α<0
Conductivity	σ = 1/ρ = nq²τ/m	Unit: (Ω·m)⁻¹ or S/m
Series resistors	R_eff = R₁+R₂+...	Same current; voltage divides
Parallel resistors	1/R_eff = 1/R₁+1/R₂+...	Same voltage; current divides
Two in parallel	R = R₁R₂/(R₁+R₂)	“Product over Sum”

📅 Section B — EMF, Kirchhoff's Laws & Power

Quantity	Formula	Key Note
Terminal voltage (discharge)	V = ε − Ir	V < ε when current is drawn
Terminal voltage (charge)	V = ε + Ir	V > ε during charging
Current in circuit	I = ε/(R+r)	r = internal resistance
Short-circuit current	I_sc = ε/r	When R = 0
Open circuit voltage	V_oc = ε	When I = 0 (no load)
Max power transfer	P_max = ε²/(4r) when R=r	R_load = r for max output
KCL (Junction Rule)	∑I_in = ∑I_out	Based on conservation of charge
KVL (Loop Rule)	∑ε = ∑IR (around any loop)	Based on conservation of energy
Power dissipated	P = VI = I²R = V²/R	All three forms are equivalent
Joule heating	H = I²Rt	Heat generated in resistance R
Energy unit	1 kWh = 3.6×10⁶ J	1 ‘unit’ of electricity = 1 kWh
Cells in series	EMF = nε; r_total = nr	Better when R >> r
Cells in parallel	EMF = ε; r_eff = r/n	Better when R << r

📅 Section C — Wheatstone Bridge & Potentiometer

Quantity	Formula	Key Note
Wheatstone balance	P/Q = R/S (or PS = QR)	I_galv = 0 at balance
Metre bridge (unknown R)	R_x = R_s × l/(100−l)	l = balancing length in cm
Potentiometer gradient	k = V/L (V/m or V/cm)	V = PD across full wire
EMF balance	ε = kl (at null deflection)	No current flows from test cell
Comparing EMFs	ε₁/ε₂ = l₁/l₂	Same k (same potentiometer wire)
Internal resistance	r = R(l₁−l₂)/l₂	l₁=open, l₂=with resistance R

🧠 Memory Tricks

“V = ε − Ir” Terminal voltage falls below EMF when discharging (energy lost in internal resistance). Rises above EMF when charging. Open circuit: V = ε exactly (I = 0, no loss).

“Series: P ∝ R | Parallel: P ∝ 1/R” In series (same I): power P = I²R → more power in larger R. In parallel (same V): P = V²/R → more power in smaller R. (Series – big R gets more power; Parallel – small R gets more power.)

“Potentiometer: No I, No Error” At balance point, zero current from test cell → no voltage drop across internal resistance → measures true EMF. Voltmeter draws small current → reads V = ε−Ir (terminal voltage only, always < EMF).

“PS = QR (Wheatstone)” Products of opposite arms are equal at balance. Remember: P/Q = R/S → cross multiply: PS = QR. Metre bridge: same principle with l and (100−l) as the two arms.

“v_d is tiny but current flows fast” Drift velocity ~10⁻⁴ m/s. Electric field signal propagates at ~c. Like water in a full pipe: push one end, other end flows instantly. Random velocity ~10⁶ m/s; drift is a small net displacement.

“Max Power when R = r” Maximum power is delivered to external load when R_load = r (internal resistance). P_max = ε²/(4r). Efficiency at this point = 50% only. Use series cells when R >> r; parallel when R << r.

🔢 Critical Values

ρ_Cu = 1.7×10⁻⁸ Ω·m ρ_Ag = 1.6×10⁻⁸ Ω·m (lowest) Relaxation time τ ~ 10⁻¹⁴ s 1 kWh = 3.6×10⁶ J Open circuit: V = ε Short circuit: I = ε/r Max power: R = r Potentiometer: I_g = 0 at balance

❌ Common Mistakes to Avoid

Resistivity vs Resistance: R = ρL/A. Resistivity ρ is a material property (independent of shape/size). Resistance R depends on shape (L and A).
Drift velocity direction: Drift of electrons is opposite to E (and opposite to conventional current direction).
KVL sign convention: EMF is +ve when traversed from − to + through the source. IR is −ve (voltage drop) in the direction of current.
Potentiometer doesn't measure terminal voltage: It measures true EMF (at null condition, I = 0, so no drop across r). Voltmeter always reads V = ε − Ir.
Parallel combination R < smallest R: Adding more resistors in parallel always decreases total resistance below that of the smallest individual resistor.
Series cells vs Parallel cells: Series → more EMF (adds up), more r (adds up). Use when external R is large. Parallel → same EMF, smaller r (divides). Use when external R is small.

📱 Practice MCQs for this topic inside our App