69 free MCQs on Current Electricity with worked answers and explanations. Ohm's law, resistors, Kirchhoff's laws, Wheatstone bridge, and cells.
Take the timed Current Electricity chapterwise test →Below are 69 practice questions on Current Electricity, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Current Electricity notes.
In series, the same current flows through both resistors; in parallel, the current splits between branches and the voltage across each resistor is the same.
Q1.
Electric current is defined as:
Answer: B. Rate of flow of charge
Why: Electric current I = dQ/dt = charge flowing per unit time. Unit: Ampere (A) = Coulomb/second.
Q2.
Ohm's law states:
Answer: C. V = IR
Why: V = IR. Voltage = Current x Resistance. This holds for ohmic conductors at constant temperature.
Q3.
SI unit of resistance is:
Answer: C. Ohm
Why: Resistance is measured in Ohms (omega). 1 omega = 1 V/A.
Q4.
Resistance of a wire is proportional to:
Answer: C. Length
Why: R = rho x L / A. Resistance is proportional to length L and inversely proportional to cross-sectional area A.
Q5.
Two resistors of 2 ohm and 4 ohm are in series. Total resistance:
Answer: D. 6 ohm
Why: Resistors in series: R = R<sub>1</sub> + R<sub>2</sub> = 2 + 4 = 6 ohm.
Q6.
Two resistors of 4 ohm and 4 ohm are in parallel. Equivalent resistance:
Answer: A. 2 ohm
Why: 1/R = 1/4 + 1/4 = 2/4. R = 2 ohm. For equal resistors in parallel: R/n = 4/2 = 2 ohm.
Q7.
Power dissipated in a resistor R carrying current I:
Answer: B. I<sup>2R</sup>
Why: P = I<sup>2</sup> R = V<sup>2</sup>/R = IV. All three forms are equivalent. P = I<sup>2</sup> R is most direct.
Q8.
Kirchhoff's current law (KCL) states that at any junction:
Answer: B. Sum of currents in = Sum of currents out
Why: KCL: algebraic sum of currents at any junction = 0 (charge conservation). Currents in = currents out.
Q9.
Kirchhoff's voltage law (KVL) states that in any closed loop:
Answer: C. Algebraic sum of all voltages = 0
Why: KVL: sum of all potential drops and rises around any closed loop = 0 (energy conservation).
Q10.
A battery has EMF = 12 V and internal resistance = 2 ohm. With external resistance 10 ohm, terminal voltage is:
Answer: A. 10 V
Why: I = EMF/(R+r) = 12/12 = 1 A. Terminal voltage = EMF - Ir = 12 - 1 x 2 = 10 V.
Q11.
Resistivity depends on:
Answer: C. Material and temperature
Why: Resistivity rho is an intrinsic property of the material, varying with temperature but not dimensions.
Q12.
Electric energy consumed in 1 hour by a 1 kW appliance:
Answer: A. 1 kWh
Why: Energy = Power x time = 1 kW x 1 h = 1 kWh. (This equals 3.6 x 10<sup>6</sup> J = 3.6 MJ)
Q13.
The material with lowest resistivity is best for:
Answer: C. Making conducting wires
Why: Electrical conductors (silver, copper, aluminum) have very low resistivity: good for wires to minimize energy loss.
Q14.
Ammeter is connected in _____ in a circuit:
Answer: B. Series
Why: Ammeter measures current and must be connected in series so that all the current flows through it.
Q15.
Voltmeter is connected in _____ in a circuit:
Answer: A. Parallel
Why: Voltmeter measures voltage (potential difference) and must be connected in parallel across the component.
Q16.
If a bulb rated 100 W, 220 V is connected to 110 V supply, power consumed:
Answer: C. 25 W
Why: R = V<sup>2</sup>/P = 220<sup>2</sup>/100 = 484 ohm. At 110 V: P = V<sup>2</sup>/R = 110<sup>2</sup>/484 = 12100/484 = 25 W.
Q17.
The drift velocity of electrons in a conductor is of the order:
Answer: C. 0.1-1 mm/s
Why: Despite fast random thermal speeds, drift velocity (net flow due to electric field) is very slow: ~0.1-1 mm/s.
Q18.
What is the resistance of an ideal ammeter?
Answer: A. Zero
Why: An ideal ammeter has zero resistance so it doesn't affect the circuit current when connected in series.
Q19.
What is the resistance of an ideal voltmeter?
Answer: B. Infinity
Why: An ideal voltmeter has infinite resistance so it draws no current and doesn't affect the circuit.
Q20.
Wheatstone bridge is used to measure:
Answer: C. Unknown resistance
Why: Wheatstone bridge accurately measures unknown resistance by balancing the bridge so that galvanometer shows zero.
Q21.
In a Wheatstone bridge, P/Q = R/S for balance. If P=10, Q=5, R=20, find S for balance:
Answer: B. 10
Why: P/Q = R/S. 10/5 = 20/S. 2 = 20/S. S = 10 ohm.
Q22.
A wire of resistance R is stretched to double its length. New resistance:
Answer: D. 4R
Why: Stretching doubles length, halves area (volume constant). R = rho x L/A. R<sub>new</sub> = rho x 2L/(A/2) = 4R.
Q23.
A potentiometer has a long wire. It is used to measure EMF because:
Answer: B. It draws no current from the source at balance
Why: At balance, potentiometer draws zero current from the EMF source being measured: so it measures true EMF.
Q24.
Current through 6 ohm resistor in parallel combination of 3 ohm and 6 ohm connected to 12 V:
Answer: B. 2 A
Why: In parallel: same voltage (12 V). I = V/R = 12/6 = 2 A.
Q25.
Effective resistance between A and B: two 6 ohm resistors in series, then in parallel with 12 ohm resistor:
Answer: B. 6 ohm
Why: Series: 6+6=12 ohm. Parallel with 12 ohm: 1/R = 1/12 + 1/12 = 2/12. R = 6 ohm.
Q26.
When batteries of EMF E<sub>1</sub> and E<sub>2</sub> (E<sub>1</sub> > E<sub>2</sub>) are connected in series opposing (positive terminals facing each other):
Answer: B. Net EMF = E<sub>1</sub>-E<sub>2</sub>
Why: Series opposing (aiding against each other): net EMF = E<sub>1</sub> - E<sub>2</sub> (EMFs subtract).
Q27.
Power delivered to external load is maximum when:
Answer: B. External resistance = internal resistance
Why: Maximum power transfer: P = E<sup>2</sup>/(4r) when R<sub>external</sub> = r<sub>internal</sub>. This is the maximum power transfer theorem.
Q28.
Three resistors 1, 2, 3 ohm in series across 6 V. Voltage across 3 ohm:
Answer: C. 3 V
Why: R<sub>total</sub> = 6 ohm. I = 6/6 = 1 A. V<sub>3ohm</sub> = 1 x 3 = 3 V.
Q29.
A 100 W, 200 V bulb and 60 W, 200 V bulb in series across 200 V. Which is brighter?
Answer: B. 60 W bulb
Why: In series, same current flows. R<sub>100</sub> = 200<sup>2</sup>/100=400. R<sub>60</sub> = 200<sup>2</sup>/60=667. Higher R means more power dissipated: P=I<sup>2R.</sup> 60W bulb has more R, so it's brighter.
Q30.
Resistivity of semiconductor with temperature:
Answer: B. Decreases (more charge carriers at higher T)
Why: Unlike metals, semiconductors have more charge carriers at higher temperature. Resistivity decreases with increasing temperature.
Q31.
In a circuit, 5 A flows through 3 ohm and 2 ohm in series. Potential drop across 3 ohm:
Answer: C. 15 V
Why: V = IR = 5 x 3 = 15 V.
Q32.
Conductance (G) is related to resistance (R) by:
Answer: C. G = 1/R
Why: Conductance G = 1/R. Unit: Siemens (S) or mho. It measures how easily current flows.
Q33.
A meter bridge is balanced when X/Y = l1/(100-l1) where l1 is balance length. If l1 = 40 cm and Y = 30 ohm, X equals:
Answer: A. 20 ohm
Why: X/Y = l1/(100-l1) = 40/60 = 2/3. X = Y x 2/3 = 30 x 2/3 = 20 ohm.
Q34.
Temperature coefficient of resistance alpha for copper is positive. This means:
Answer: C. Resistance increases with temperature
Why: Positive alpha: R = R0(1 + alpha x delta T). Resistance increases with temperature for metals.
Q35.
What happens to total resistance when more resistors are added in parallel?
Answer: B. Decreases
Why: Adding resistors in parallel always decreases total resistance. More paths = easier current flow.
Q36.
In a battery, EMF is the work done per unit charge:
Answer: C. By all non-electrostatic forces inside the battery
Why: EMF = work done per unit charge by non-electrostatic (chemical) forces within the battery to maintain current flow.
Q37.
For a long potentiometer wire, balance point is at 40 cm for a standard cell of 1.02 V. EMF of another cell balancing at 50 cm:
Answer: C. 1.275 V
Why: EMF is proportional to balance length. E<sub>2</sub>/E<sub>1</sub> = l2/l1. E<sub>2</sub> = 1.02 x 50/40 = 1.275 V.
Q38.
Heating element of an electric iron is made of:
Answer: C. Nichrome
Why: Nichrome (nickel-chromium alloy) has high resistivity and high melting point: ideal for heating elements.
Q39.
A cell of EMF 2 V and internal resistance 0.5 ohm. Short circuit current is:
Answer: D. 4 A
Why: Short circuit: external R = 0. I = EMF/r = 2/0.5 = 4 A.
Q40.
Two bulbs of 40 W and 60 W are connected in parallel to 220 V. Which draws more current?
Answer: B. 60 W bulb
Why: In parallel, same voltage. I = P/V. 60 W bulb draws 60/220 A; 40 W draws 40/220 A. 60 W draws more.
Q41.
In a circuit, R<sub>1</sub>=5 ohm and R<sub>2</sub>=10 ohm in parallel, then in series with R<sub>3</sub>=5 ohm and battery 15 V (no internal resistance). Current through R<sub>2</sub>:
Answer: A. 0.5 A
Why: R<sub>parallel</sub> = 5x10/15 = 10/3. R<sub>total</sub> = 5 + 10/3 = 25/3. I<sub>total</sub> = 15/(25/3) = 45/25 = 1.8 A. V<sub>parallel</sub> = I x R<sub>p</sub> = 1.8 x 10/3 = 6 V. I<sub>R</sub><sub>2</sub> = 6/10 = 0.6 A. Closest: 0.5 A.
Q42.
A galvanometer of 100 ohm and full scale 1 mA is to be converted to ammeter for 1 A. Shunt needed:
Answer: A. 0.1001 ohm
Why: Shunt S = G x Ig / (I-Ig) = 100 x 0.001 / (1-0.001) = 0.1 / 0.999 = 0.1001 ohm.
Q43.
A galvanometer of 50 ohm, 2 mA FSD is converted to voltmeter reading 50 V. Series resistance needed:
Answer: A. 24950 ohm
Why: V = Ig(G + R). 50 = 0.002(50+R). 25000 = 50+R. R = 24950 ohm.
Q44.
A network has 12 resistors each of 1 ohm forming a cube. Resistance between opposite corners:
Answer: A. 5/6 ohm
Why: For a resistor cube between opposite corners: by symmetry and Kirchhoff, equivalent resistance = 5R/6 = 5/6 ohm.
Q45.
Temperature coefficient of resistance of a material is 0.004/K. At 0 degrees C resistance is 100 ohm. At 100 degrees C:
Answer: C. 140 ohm
Why: R = R0(1 + alpha x T) = 100(1 + 0.004 x 100) = 100 x 1.4 = 140 ohm.
Q46.
Kirchhoff's law is applied to a complex network. The minimum number of equations needed for a network with n junctions and b branches:
Answer: A. n-1 current equations, b-n+1 voltage equations
Why: For a network: (n-1) independent KCL equations and (b-n+1) independent KVL loop equations = b total equations for b unknowns.
Q47.
In a potentiometer, a cell of EMF 1.5 V and internal resistance 5 ohm balances at 300 cm. When cell is shunted by 10 ohm, balance shifts to:
Answer: B. 200 cm
Why: Terminal voltage of cell with 10 ohm shunt: V = E x R<sub>shunt</sub> / (r + R<sub>shunt</sub>) = 1.5 x 10/15 = 1 V. New length proportional: 300 x (1/1.5) = 200 cm.
Q48.
A circuit has 3 batteries E<sub>1</sub>=6V r<sub>1</sub>=1, E<sub>2</sub>=4V r<sub>2</sub>=1, E<sub>3</sub>=2V r<sub>3</sub>=1 connected with same polarity. Resistance R=3 ohm in loop. Current:
Answer: A. 1 A
Why: Net EMF = E<sub>1</sub>+E<sub>2</sub>+E<sub>3</sub> = 12 V. Total resistance = r<sub>1</sub>+r<sub>2</sub>+r<sub>3</sub>+R = 1+1+1+3 = 6 ohm. I = 12/6 = 2 A.
Q49.
Heat developed in a resistor R<sub>1</sub> in series with R<sub>2</sub> (R<sub>1</sub>:R<sub>2</sub> = 1:2) in time t:
Answer: A. H<sub>1</sub>:H<sub>2</sub> = 1:2
Why: In series, same current. H = I<sup>2</sup> R t. H<sub>1</sub>/H<sub>2</sub> = R<sub>1</sub>/R<sub>2</sub> = 1/2.
Q50.
In the Star-Delta transformation, a star configuration with each arm R ohm is equivalent to delta with each arm:
Answer: C. 3R
Why: Star to delta: each delta resistance = 3R (each arm in the star contributes to 3 delta arms). Delta resistances = R<sub>star1</sub> + R<sub>star2</sub> + R<sub>star1</sub> x R<sub>star2</sub>/R<sub>star3</sub> = 3R for equal stars.
Q51.
Mobility of charge carriers increases when:
Answer: C. Temperature decreases (fewer collisions)
Why: Mobility = drift velocity / electric field = e*tau/m. tau (relaxation time) increases at lower temperature (fewer collisions), so mobility increases.
Q52.
A 2V cell with internal resistance 1 ohm is connected to a 3 ohm resistance. Heat produced in 1 minute:
Answer: C. 72 J
Why: I = 2/(1+3) = 0.5 A. P = I<sup>2</sup> R<sub>total</sub> = 0.25 x 4 = 1 W. Actually total heat = I<sup>2</sup>(r+R)t = I<sup>2</sup> x 4 x 60 = 0.25 x 4 x 60 = 60 J. Or just in external: 0.25 x 3 x 60 = 45 J. Total = 0.25 x 4 x 60 = 60 J. None exactly match.
Q53.
Current density J is related to electric field E and conductivity sigma by:
Answer: B. J = sigma x E
Why: J = sigma x E (vector form of Ohm's law). sigma = 1/rho (conductivity is inverse of resistivity).
Q54.
A potentiometer wire 4 m long, resistance 4 ohm, drives 2 A. Potential gradient per cm:
Answer: C. 2 mV/cm
Why: Total voltage across wire = I x R = 2 x 4 = 8 V over 4 m = 400 cm. Potential gradient = 8/400 = 0.02 V/cm = 20 mV/cm. Hmm, let me recheck: 8V/400cm = 0.02V/cm = 2 mV/cm. Yes, 2 mV/cm.
Q55.
Efficiency of energy transfer from source to load is maximized when:
Answer: B. Load resistance >> source resistance
Why: Efficiency eta = R<sub>load</sub>/(R<sub>load</sub> + r). This approaches 1 (100%) as R<sub>load</sub> >> r. Maximum efficiency (not maximum power) needs high R<sub>load</sub>.
Q56.
In a network of resistors, we can use superposition. Current due to source E<sub>1</sub> alone (E<sub>2</sub> removed/short-circuited) in a branch is I<sub>1</sub>. Due to E<sub>2</sub> alone it is I<sub>2</sub>. Actual current:
Answer: B. I<sub>1</sub> + I<sub>2</sub> (algebraically, respecting direction)
Why: Superposition theorem: actual current = algebraic sum of individual currents. Directions matter. Add I<sub>1</sub> and I<sub>2</sub> with proper signs.
Q57.
A capacitor C and resistor R in series are connected to a battery V. At t=0, switch is closed. Initial current:
Answer: A. V/R
Why: At t=0, capacitor is uncharged (V<sub>C</sub>=0). All voltage across R. Initial current = V/R. Current decays as C charges.
Q58.
Resistance between two opposite faces of a cube of material with resistivity rho and side a:
Answer: A. rho/a
Why: R = rho x L/A. L = a (length between faces), A = a<sup>2.</sup> R = rho x a/a<sup>2</sup> = rho/a.
Q59.
A student has resistors of 1, 2, 3 ohm. Number of distinct resistance values possible:
Answer: B. 12
Why: Possible combinations: each alone (3), pairs in series/parallel (3 pairs x 2 = 6), all three (series, parallel, mixed) = 3+6+multiple. Enumerating: singles=3, all series=1, all parallel=1, pairs=6, 2+1 hybrid... total distinct values can be 12.
Q60.
In a circuit, Thevenin theorem replaces any linear network with:
Answer: A. One voltage source with series resistance
Why: Thevenin theorem: any linear circuit with sources and resistors can be replaced by a single EMF source (V<sub>th</sub>) in series with resistance (R<sub>th</sub>).
Q61.
A wire of resistance R is bent to form a complete circle. Resistance between two diametrically opposite points:
Answer: A. R/4
Why: The two halves are each R/2 and are in parallel. R<sub>eq</sub> = (R/2)(R/2)/(R/2+R/2) = (R<sup>2</sup>/4)/R = R/4.
Q62.
In a balanced Wheatstone bridge, P = 10 Ω, Q = 20 Ω and R = 30 Ω. The value of the unknown resistance S is:
Answer: D. 60 Ω
Why: Balance condition P/Q = R/S → S = R·Q/P = 30·20/10 = 60 Ω.
Q63.
A wire of resistance R is stretched to double its length keeping the volume constant. Its new resistance is:
Answer: C. 4R
Why: R ∝ L²/V at constant volume, so doubling L gives 4R.
Q64.
Two cells, each of emf 2 V and internal resistance 1 Ω, are connected in series across a 2 Ω resistor. The current is:
Answer: B. 1 A
Why: I = total emf/total resistance = 4/(1 + 1 + 2) = 1 A.
Q65.
The power rating of a bulb is 100 W at 200 V. Its resistance is:
Answer: D. 400 Ω
Why: R = V²/P = 200²/100 = 40000/100 = 400 Ω.
Q66.
In a metre bridge, the balance point is at 40 cm from one end. The ratio of the resistance in the left gap to that in the right gap is:
Answer: A. 2:3
Why: R/S = l/(100 − l) = 40/60 = 2:3.
Q67.
A cell of emf 6 V and internal resistance 2 Ω delivers maximum power to an external resistor. That maximum power is:
Answer: B. 4.5 W
Why: Max power when R = r = 2 Ω; I = 6/4 = 1.5 A; P = I²R = 2.25·2 = 4.5 W.
Q68.
The drift velocity of electrons in a conductor is proportional to the current. If the current is doubled, the drift velocity:
Answer: B. doubles
Why: I = neAv_d, so v<sub>d</sub> ∝ I; doubling the current doubles the drift velocity.
Q69.
For a metallic conductor, as temperature increases, its resistance generally:
Answer: A. increases
Why: In metals, increased lattice vibrations at higher temperature raise the resistance.