SUMMARY: The chapter on Current Electricity in Class 12 Physics focuses on the study of electric current, its flow in conductors, and the laws governing it. KEY TOPICS: Electric current, Ohm's law, resistivity, electrical energy and power, series and parallel combinations of resistors, Kirchhoff's laws, Wheatstone bridge, potentiometer, internal resistance of a cell.
Ohm's law states that V = IR. The SI unit of resistance R is:
ACoulomb
BVolt
COhm (Ω)
DWatt
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Correct answer: Option 3 — Ohm (Ω)
Q31 Mark
Two resistors of 4 Ω and 6 Ω in series have an equivalent resistance of:
A2.4 Ω
B5 Ω
C8 Ω
D10 Ω
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Correct answer: Option 4 — 10 Ω
Q41 Mark
The drift velocity of electrons is in the order of:
A10⁸ m/s
B10⁵ m/s
C10⁻⁴ m/s
D10⁻¹⁰ m/s
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Correct answer: Option 3 — 10⁻⁴ m/s
Q51 Mark
Power dissipated in a resistor R carrying current I is:
AIR
BIR²
CI²R
DIR³
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Correct answer: Option 3 — I²R
Short Answer Questions5 questions
Q63 Marks
State Ohm's law and write its mathematical form.
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Ohm's law: at constant temperature the current flowing through a conductor is directly proportional to the potential difference applied across it. V = IR where V is potential difference (V) I is current (A) and R is resistance (Ω). The law is empirical and applies to most metallic conductors in normal conditions; it doesn't apply to non-ohmic devices like diodes or semiconductors.
Q73 Marks
Define drift velocity. How is it related to current?
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Drift velocity v_d is the average velocity acquired by free electrons in a conductor under the influence of an applied electric field. It is much smaller than the random thermal velocity (~10⁵ m/s) but is responsible for net current flow. Relation to current: I = nAev_d where n is the number density of free electrons A is the cross-sectional area e is electronic charge.
Q83 Marks
Find the equivalent resistance of three resistors 4 Ω 6 Ω 12 Ω connected in parallel.
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For parallel resistors: 1/R_eq = 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12 = 1/2 ⇒ R_eq = 2 Ω. The equivalent resistance is always less than the smallest individual resistance.
Q93 Marks
State Kirchhoff's two laws.
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Kirchhoff's current law (KCL or junction rule): the algebraic sum of currents entering any junction equals the sum leaving — Σ I_in = Σ I_out. Based on conservation of charge. Kirchhoff's voltage law (KVL or loop rule): the algebraic sum of EMFs around any closed loop equals the sum of IR drops — Σ E = Σ IR. Based on conservation of energy. Together they allow analysis of any DC circuit.
Q103 Marks
Differentiate between EMF and terminal voltage.
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EMF (electromotive force) E is the maximum potential difference a source can provide when no current is drawn (open-circuit voltage). Terminal voltage V is the actual potential difference across the terminals when current I flows: V = E − Ir where r is the internal resistance of the source. So V < E whenever current flows; V = E only at zero current. For a discharging battery V decreases as I increases.
Long Answer Questions6 questions
Q116 Marks
Derive the equivalent resistance for resistors connected in series and parallel.
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Series: same current I flows through each resistor; total voltage = sum of individual drops: V = V₁ + V₂ + V₃ = IR₁ + IR₂ + IR₃ = I(R₁ + R₂ + R₃). So R_eq = R₁ + R₂ + R₃ — series resistances ADD. Parallel: same voltage V across each resistor; total current = sum: I = I₁ + I₂ + I₃ = V/R₁ + V/R₂ + V/R₃ = V(1/R₁ + 1/R₂ + 1/R₃). So 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ — RECIPROCALS add in parallel.
Q126 Marks
In a circuit a 12 V battery with internal resistance 1 Ω is connected to a resistor of 5 Ω. Compute (i) current in the circuit, (ii) terminal voltage of the battery, (iii) power dissipated in the external resistor.
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Total resistance: R_total = R_external + r = 5 + 1 = 6 Ω. (i) Current: I = E/R_total = 12/6 = 2 A. (ii) Terminal voltage: V = E − Ir = 12 − 2(1) = 10 V. (or V = IR = 2 × 5 = 10 V). (iii) Power in external resistor: P = I²R = 4 × 5 = 20 W. Power lost in internal resistance: P_int = I²r = 4 × 1 = 4 W. Total power supplied by battery: 24 W = EI = 12 × 2 ✓.
Q136 Marks
Derive the expression for Wheatstone bridge balance condition.
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A Wheatstone bridge has four resistors P Q R S in a diamond arrangement with a galvanometer between B and D. Let battery be connected between A and C. Apply Kirchhoff's laws: at junction B: I₁ = I_g + I₂. At balance galvanometer current I_g = 0 ⇒ I₁ = I₂ and I₃ = I₄ (similar reasoning for D). Voltage drop: V_AB = V_AD ⇒ I₁P = I₃R; V_BC = V_DC ⇒ I₂Q = I₄S. Dividing: P/Q = R/S — the balance condition. Used to measure unknown resistances precisely by balancing against known ones.
Q146 Marks
Discuss the variation of resistance of a metallic conductor with temperature.
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For most metallic conductors resistance increases linearly with temperature for moderate ranges: R_t = R_0[1 + α(t − t_0)] where α is the temperature coefficient of resistance (typically ~0.004 /°C for copper). Cause: at higher temperatures lattice ions vibrate more energetically increasing the frequency of electron-ion collisions and hence reducing the mean free path and drift velocity — increasing resistance. At very low temperatures resistance approaches a residual value (impurity-dependent) and some materials become superconducting (R = 0). Semiconductors show the opposite behaviour: R decreases with T because more charge carriers are thermally activated.
Q156 Marks
A wire of resistance 4 Ω is bent into the form of a circle and a current of 6 A enters at one diametrically opposite end and leaves at the other. Find the current in each semicircle and the equivalent resistance of the loop.
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The wire has total resistance 4 Ω so each semicircle has resistance 2 Ω (half the wire). The two semicircles are in parallel between the entry and exit points: R_eq = (2 × 2)/(2 + 2) = 1 Ω. Total current 6 A splits equally between equal-resistance semicircles: I₁ = I₂ = 3 A. Voltage between entry and exit: V = IR_eq = 6 × 1 = 6 V. Current through each: I = V/R_each = 6/2 = 3 A ✓.
Q166 Marks
Differentiate between EMF and potential difference in tabular form on five features.
Assertion–Reason Questions5 questions
Q171 Mark
Assertion (A): Ohm's law applies to metallic conductors at constant temperature.
Reason (R): At higher temperatures the resistance changes so V = IR is not strictly linear without temperature control.
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Correct answer: Option 1 —
Both A and R are true, and R is the correct explanation of A.
Q181 Mark
Assertion (A): Drift velocity of electrons is much smaller than their thermal velocity.
Reason (R): Random thermal motion is dominant; the applied field produces only a small bias which is the drift velocity.
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Correct answer: Option 1 —
Both A and R are true, and R is the correct explanation of A.
Q191 Mark
Assertion (A): Kirchhoff's current law is a consequence of conservation of charge.
Reason (R): At a junction the rate at which charge enters must equal the rate at which it leaves — no charge can accumulate in a wire.
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Correct answer: Option 1 —
Both A and R are true, and R is the correct explanation of A.
Q201 Mark
Assertion (A): Power dissipated as heat in a resistor is P = I²R = V²/R.
Reason (R): Both forms are equivalent using Ohm's law V = IR; heat is generated by collisions of charge carriers with the lattice.
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Correct answer: Option 1 —
Both A and R are true, and R is the correct explanation of A.
Q211 Mark
Assertion (A): A superconductor offers zero resistance to current flow.
Reason (R): Below its critical temperature electron pairs (Cooper pairs) move through the lattice without scattering — no energy is dissipated.
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Correct answer: Option 1 —
Both A and R are true, and R is the correct explanation of A.
Statement-Based Questions5 questions
Q221 Mark
Statement 1: Current is the rate of flow of charge.
Statement 2: The SI unit of current is the ampere.
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Correct answer: Option 1 —
Both statements are true.
Q231 Mark
Statement 1: V = IR holds at constant temperature.
Statement 2: Resistance has SI unit ohm (Ω).
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Correct answer: Option 1 —
Both statements are true.
Q241 Mark
Statement 1: Resistors in series add: R_eq = ΣR_i.
Correct answer: Option 1 —
Both statements are true.
Q251 Mark
Statement 1: Kirchhoff's current law: at any junction Σ currents in = Σ currents out.
Statement 2: Kirchhoff's voltage law: around any closed loop Σ EMF = Σ IR.
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Correct answer: Option 1 —
Both statements are true.
Q261 Mark
Statement 1: The terminal voltage of a discharging cell is less than its EMF.
Statement 2: V = E − Ir where r is internal resistance.
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Correct answer: Option 1 —
Both statements are true.
Case Study / Passage Questions3 questions
Q273 Marks
In a Wheatstone bridge P = 10 Ω Q = 20 Ω R = 30 Ω. The student wants to find the unknown resistance S for the bridge to be balanced (galvanometer reading zero) and verify the principle.
The unknown resistance S for balance equals:
A15 Ω
B30 Ω
C60 Ω
D90 Ω
The balance condition for a Wheatstone bridge is:
AP×Q = R×S
BP/Q = R/S
CP/R = Q/S
DP×R = Q×S
Why does the Wheatstone bridge give zero galvanometer reading at balance?
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1. Option 3 — 60 Ω
2. Option 2 — P/Q = R/S
3. At balance no current flows through the galvanometer. The condition is P/Q = R/S ⇒ S = QR/P = (20 × 30)/10 = 60 Ω. Equivalently P × S = Q × R gives the same result. The Wheatstone bridge is used as a precise method to determine an unknown resistance by adjusting two known resistances until the galvanometer shows zero. The Meter Bridge is a practical realization on a slide-wire.
Q283 Marks
A 12 V battery is connected to a load. When no current is drawn the voltmeter reads 12 V (open-circuit voltage). When 4 A flows the voltmeter reads 10 V. The mechanic wants to find the internal resistance of the battery.
The internal resistance of the battery equals:
A0.5 Ω
B1 Ω
C2 Ω
D4 Ω
The difference of 12 V − 10 V = 2 V is due to:
AOpen-circuit voltage
BClosed-circuit current
CEMF
DVoltage drop on internal resistance
Compute the power dissipated in the internal resistance when 4 A flows.
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1. Option 1 — 0.5 Ω
2. Option 4 — Voltage drop on internal resistance
3. EMF: E = open-circuit voltage = 12 V (when I = 0). Terminal voltage: V = E − Ir. With I = 4 A V = 10: 10 = 12 − 4r ⇒ r = 0.5 Ω. The internal resistance dissipates power even when the external load is varied. As the battery ages r increases — eventually the battery cannot supply useful current. Old/deeply discharged batteries can have r > 1 Ω which is why they show low terminal voltages under load even when 'fully charged'.
Q293 Marks
A circuit contains a 12 V battery in series with a 2 Ω resistor and then a parallel combination of 6 Ω and 12 Ω resistors. The student wants to find the total resistance the current from the battery and the current through each parallel branch.
The equivalent resistance of the entire circuit equals:
A2 Ω
B4 Ω
C6 Ω
D8 Ω
The total current from the battery equals:
A1 A
B2 A
C3 A
D4 A
Verify the sum of currents through each parallel branch.
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1. Option 3 — 6 Ω
2. Option 2 — 2 A
3. Parallel combination of 6 Ω and 12 Ω: 1/R_p = 1/6 + 1/12 = 3/12 = 1/4 ⇒ R_p = 4 Ω. Total: R_total = R_series + R_p = 2 + 4 = 6 Ω. Current from battery: I = V/R_total = 12/6 = 2 A. Voltage across parallel section = 2 × 4 = 8 V. Current through 6 Ω: I₁ = 8/6 = 1.33 A. Current through 12 Ω: I₂ = 8/12 = 0.67 A. Total: 1.33 + 0.67 = 2 A ✓.
Table-Based Questions4 questions
Q303 Marks
Study Ohm's law and power formulas:
Quantity
Formula
Voltage
V = IR
Current
I = V/R
Resistance
R = V/I
Power (DC)
P = VI = I²R = V²/R
Energy
W = Pt = VIt
The power dissipated in a resistor R carrying current I across voltage V can be expressed as:
AVI
BI²R
CV²/R
DAll three
Power varies as ___ with current at fixed R:
ALinear
BInverse
CQuadratic
DLogarithmic
Calculate the power dissipated when 5 A flows through a 4 Ω resistor.
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1. Option 4 — All three
2. Option 3 — Quadratic
3. Three equivalent forms of power: P = VI (general); P = I²R (when current is known and you want to find heating); P = V²/R (when voltage is fixed across the resistor). All follow from Ohm's law V = IR. Power dissipation produces heat (Joule heating) which is the basis of incandescent bulbs heaters and toasters. In long-distance transmission high-voltage low-current is used to minimize I²R losses.
Q313 Marks
Compare types of cells:
Cell type
EMF (V)
Internal resistance
Daniell cell (Zn-Cu)
1.10
Moderate
Lechlanche dry cell
1.5
Low to moderate
Lead-acid (rechargeable)
2.0 per cell
Very low
Lithium-ion (rechargeable)
3.7
Very low
Which cell has the highest EMF per cell?
ADaniell
BLechlanche
CLead-acid
DLithium-ion
Which cells are rechargeable?
ALithium-ion only
BLead-acid only
CBoth A and B
DLechlanche
Which cell would you choose for a smartphone — and why?
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1. Option 4 — Lithium-ion
2. Option 3 — Both A and B
3. Different cells have different EMFs internal resistances cycle lives and energy densities. Daniell cell is historic but used in physics labs (instructive). Lechlanche dry cell is the common alkaline AA — single use. Lead-acid is heavy but very high current-output — used in cars to start engines. Lithium-ion is the modern champion — high energy density rechargeable used in phones laptops EVs. New chemistries (sodium-ion solid-state) are emerging.
Q326 Marks
In a Wheatstone bridge, P = 10 Ω, Q = 20 Ω, R = 30 Ω. Compute (i) the value of S for balance, (ii) what happens if S is increased to 80 Ω, (iii) the current through the galvanometer if S = 80 Ω, P = 10 Ω, Q = 20 Ω, R = 30 Ω, and battery EMF = 6 V (G = 100 Ω, ignore battery resistance).
Resistor
Value
P
10 Ω
Q
20 Ω
R
30 Ω
Q336 Marks
A cell of EMF 12 V and internal resistance 1 Ω is connected to an external resistance R. Compute (i) terminal voltage when R = 3 Ω, (ii) current and power dissipated when R = 3 Ω, (iii) the value of R for maximum power transfer and the maximum power.
Quantity
Symbol
Value
EMF
ε
12 V
Internal resistance
r
1 Ω
External resistance
R
3 Ω
Picture-Based Questions1 question
Q343 Marks
Study the Wheatstone bridge circuit and answer:
The condition for the Wheatstone bridge to be balanced is:
AP/Q = R/S
BP × Q = R × S
CP + Q = R + S
DP − Q = R − S
At balance, the current through the galvanometer is:
AMaximum
BZero
CVariable
DNegative
Derive the balance condition and explain its significance.
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1. Option 1 — P/Q = R/S
2. Option 2 — Zero
3. The Wheatstone bridge consists of four resistances P, Q, R, S in a diamond, with a galvanometer between two opposite vertices and a battery between the other two. At balance the galvanometer shows zero deflection: this means the potential at B equals that at D, which gives P/Q = R/S. The bridge is sensitive — a small change in any resistance produces a measurable galvanometer current. Used in resistance measurement (using known P, Q and comparing R with unknown S) and in strain gauges where mechanical strain alters resistance.
