Question 1

A wire of resistance 10 Ω carries a current of 2 A. What is the potential difference across the wire?

Accepted Answer

Using Ohm's law, V = IR = 2 × 10 = 20 V.

Question 2

The SI unit of electrical resistivity is:

Accepted Answer

Resistivity ρ = RA/l, where R is in ohms, A in m², and l in m; thus unit is Ω·m.

Question 3

Two resistors of 6 Ω and 12 Ω are connected in parallel. What is the equivalent resistance?

Accepted Answer

For parallel resistors: 1/R_eq = 1/6 + 1/12 = 3/12 = 1/4, so R_eq = 4 Ω.

Question 4

A conductor has a length of 2 m, cross-sectional area of 1 mm², and resistivity of 1.6 × 10⁻⁸ Ω·m. Calculate its resistance.

Accepted Answer

R = ρl/A = (1.6 × 10⁻⁸ × 2)/(1 × 10⁻⁶) = 32 Ω.

Question 5

If the current through a resistor is doubled while keeping resistance constant, how does the power dissipated change?

Accepted Answer

Power P = I²R; if I is doubled, P becomes (2I)²R = 4I²R, which is 4 times the original.

Question 6

Three resistors of 5 Ω, 10 Ω, and 15 Ω are connected in series across a 60 V battery. What is the total current in the circuit?

Accepted Answer

Total resistance = 5 + 10 + 15 = 30 Ω; Current I = V/R = 60/30 = 2 A.

Question 7

A wire of resistance 10 Ω carries a current of 2 A. What is the potential difference across the wire?

Accepted Answer

Using Ohm's law, V = IR = 2 × 10 = 20 V.

Question 8

The resistivity of a conductor depends on which of the following factors?

Accepted Answer

Resistivity is an intrinsic property of a material that depends only on the material type and temperature, not on geometry or applied voltage.

Question 9

A copper wire of length 2 m and cross-sectional area 1 mm² has resistivity 1.6 × 10⁻⁸ Ω·m. Calculate its resistance.

Accepted Answer

R = ρL/A = (1.6 × 10⁻⁸ × 2)/(1 × 10⁻⁶) = 32 Ω.

Question 10

A wire has resistance R at 20°C. If its resistance at 100°C is 1.2R, what is the temperature coefficient of resistance?

Accepted Answer

Using R_t = R₀(1 + αΔT), 1.2R = R(1 + α × 80), so α = 0.2/80 = 2.5 × 10⁻³ °C⁻¹.

Question 11

A battery of EMF 12 V and internal resistance 2 Ω is connected to an external resistance of 4 Ω. What is the terminal voltage of the battery?

Accepted Answer

Current I = EMF/(R + r) = 12/6 = 2 A; Terminal voltage V = EMF - Ir = 12 - 2(2) = 8 V.

Question 12

According to Kirchhoff's current law, at a junction in a circuit, what must be true?

Accepted Answer

Kirchhoff's current law (conservation of charge) states that the total current entering a junction must equal the total current leaving it.

Question 13

A heating element draws a current of 5 A at 220 V. What is the power dissipated as heat?

Accepted Answer

Power P = VI = 220 × 5 = 1100 W, or P = V²/R or I²R can also be used; the answer is 1100 W.

Question 14

Two resistors of 20 Ω and 30 Ω are connected in parallel. What is their equivalent resistance?

Accepted Answer

For parallel connection: 1/R_eq = 1/20 + 1/30 = 5/60, so R_eq = 12 Ω.

Question 15

Three resistors of 6 Ω, 4 Ω, and 12 Ω are connected in parallel across a 24 V source. What is the total current drawn from the source?

Accepted Answer

For parallel resistors: 1/R_eq = 1/6 + 1/4 + 1/12 = 2/12 + 3/12 + 1/12 = 6/12 = 1/2, so R_eq = 2 Ω; I = V/R_eq = 24/2 = 12 A. Wait, recalculate: 1/R_eq = (2+3+1)/12 = 1/2, R_eq = 2 Ω is wrong. Actually 1/6 + 1/4 + 1/12 = (2+3+1)/12 = 6/12 = 1/2... R_eq should be 8/3 Ω. Let me verify: I_total = 24/6 + 24/4 + 24/12 = 4 + 6 + 2 = 12 A. Answer is C, not B.

Question 16

A heater element draws 5 A current when connected to a 220 V supply. If the element is cut into two equal pieces and both are connected in parallel to the same supply, what is the new current drawn?

Accepted Answer

Original resistance R = V/I = 220/5 = 44 Ω; each piece has resistance R/2 = 22 Ω; parallel combination gives 1/R_eq = 1/22 + 1/22 = 2/22, so R_eq = 11 Ω; new current I = 220/11 = 20 A.

Question 17

The resistivity of a conductor is 2 × 10⁻⁶ Ω·m. A wire made from this material has length 5 m and diameter 1 mm. What is its resistance?

Accepted Answer

R = ρL/A where A = π(d/2)² = π(0.5×10⁻³)² = 7.854×10⁻⁷ m²; R = (2×10⁻⁶ × 5)/(7.854×10⁻⁷) ≈ 12.74 Ω... recalculate: R = 10×10⁻⁶/7.854×10⁻⁷ ≈ 12.74 Ω, answer should be B. Using A = πr² = π(0.5mm)² ≈ 0.785 mm²: R = (2×10⁻⁶ × 5)/(0.785×10⁻⁶) ≈ 12.74 Ω.

Question 18

A wire of resistance 10 Ω is stretched to double its original length. If the resistivity and density remain constant, what is the new resistance?

Accepted Answer

Resistance R = ρL/A; when length doubles and volume remains constant, cross-sectional area becomes A/2, so R_new = ρ(2L)/(A/2) = 4ρL/A = 4R₀ = 40 Ω.

Agniveer Army Technical Current Electricity

Current Electricity — Practice Questions