Electricity — Set 1

Correct Answer: D. Coulomb

• **Coulomb** = Electric charge is measured in coulomb in the SI system. • **Q = It — 1 C = 1 A × 1 s** — One coulomb equals the charge transported by a 1 A current in 1 second. • 💡 Wrong-option analysis: Ampere: unit of current, not charge; Volt: unit of electric potential; Ohm: unit of resistance.

Correct Answer: A. V = IR

• **V = IR** = Ohm's law states that voltage across a conductor equals current times resistance. • **V = IR — doubling I doubles V for the same R** — Valid at constant temperature for ohmic conductors. • 💡 Wrong-option analysis: V = I/R: inverts the role of R giving wrong magnitude; R = VI: rearranges incorrectly making R proportional to V×I; I = VR: wrong rearrangement, I would grow with R.

Correct Answer: A. 4 Ω

• **4 Ω** = Using Ohm's law R = V/I = 12/3 = 4 Ω. • **R = V/I = 12/3 = 4 Ω** — Doubling V with the same I would give R = 8 Ω. • 💡 Wrong-option analysis: 0.25 Ω: that is I/V not V/I; 36 Ω: product V×I, not the ratio; 9 Ω: no standard arithmetic route gives this.

Correct Answer: D. 5 Ω

• **5 Ω** = In series, resistances add directly: Req = 2 + 3 = 5 Ω. • **Req = R1 + R2 = 2 + 3 = 5 Ω** — The same current flows through both, so the total voltage drop equals the sum of individual drops. • 💡 Wrong-option analysis: 6 Ω: product 2×3, not the series sum; 1.2 Ω: parallel formula result (R1R2)/(R1+R2); 0.5 Ω: no valid formula gives this.

Correct Answer: A. Req = (R1R2)/(R1 + R2)

• **Req = (R1R2)/(R1 + R2)** = For two parallel resistors, 1/Req = 1/R1 + 1/R2 simplifies to Req = R1R2/(R1+R2). • **1/Req = 1/R1 + 1/R2; Req < smallest branch** — Parallel combination always gives a resistance smaller than either branch. • 💡 Wrong-option analysis: Req = R1+R2: that is the series formula; Req = (R1+R2)/2: arithmetic mean, has no basis in circuit law; Req = R1R2: dimensionally gives Ω², not Ω.

Correct Answer: B. 4 A

• **4 A** = Current is charge per unit time: I = Q/t = 12/3 = 4 A. • **I = Q/t = 12/3 = 4 A** — Larger charge in the same time means proportionally larger current. • 💡 Wrong-option analysis: 1 A: incorrectly computes t/Q; 9 A: no arithmetic path leads here; 36 A: product Q×t instead of ratio.

Correct Answer: C. P = VI

• **P = VI** = Electric power equals voltage multiplied by current, P = VI. • **P = VI = I²R = V²/R — all three forms equivalent via Ohm's law** — Power unit is watt (W = J/s). • 💡 Wrong-option analysis: P = R/I: dimensionally gives Ω/A, not watt; P = IR: this equals voltage V, not power; P = V/I: this equals resistance R, not power.

Correct Answer: C. 0.2 kWh

• **0.2 kWh** = Energy = Power × Time = 0.1 kW × 2 h = 0.2 kWh. • **100 W = 0.1 kW; 0.1 × 2 = 0.2 kWh** — A 1000 W device running for 1 hour uses exactly 1 kWh. • 💡 Wrong-option analysis: 2 kWh: forgets to convert 100 W to 0.1 kW; 0.02 kWh: divides instead of multiplying; 20 kWh: multiplies 100 W×2 without kW conversion.

Correct Answer: B. Ω m

• **Ω m** = From R = ρL/A, resistivity ρ = RA/L gives units of Ω × m²/m = Ω m. • **ρ = RA/L → Ω·m; copper ρ ≈ 1.7×10⁻⁸ Ω m** — Insulators like rubber have ρ ≈ 10¹³ Ω m. • 💡 Wrong-option analysis: Ω/m: resistance per unit length, not resistivity; Ω m²: includes an extra metre from incorrect algebra; m/Ω: this is a unit of conductance-length, not resistivity.

Correct Answer: C. 1 Ω

• **1 Ω** = Using R = ρL/A = (2×10⁻⁷ × 5)/(1×10⁻⁶) = 10⁻⁶/10⁻⁶ = 1 Ω. • **R = (2×10⁻⁷×5)/10⁻⁶ = 10⁻⁶/10⁻⁶ = 1 Ω** — Doubling length would give 2 Ω; doubling area would give 0.5 Ω. • 💡 Wrong-option analysis: 0.01 Ω: power-of-ten error; 0.1 Ω: arithmetic slip in exponents; 2 Ω: uses L without dividing by A.