Circuits — Set 1
Physics · परिपथ · Questions 1–10 of 50
Two resistors of 2 Ω and 3 Ω are connected in series. What is the equivalent resistance?
Correct Answer: A. 5 Ω
• **5 Ω** = In series, resistances add directly: R = 2 + 3 = 5 Ω. • **R = 2 + 3 = 5 Ω** — Series connection always increases total resistance; same current flows through each resistor. • 💡 Wrong-option analysis: 2/5 Ω: that is the parallel equivalent, not series; 6 Ω: product 2×3 gives 6, not the rule for series; 1 Ω: no arithmetic route from 2 and 3 gives 1 in series.
Two resistors of 6 Ω and 3 Ω are connected in parallel. What is the equivalent resistance?
Correct Answer: A. 2 Ω
• **2 Ω** = For parallel resistors 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R = 2 Ω. • **1/R = 1/6 + 2/6 = 1/2 → R = 2 Ω** — Parallel resistance is always less than the smallest branch (3 Ω here). • 💡 Wrong-option analysis: 1/2 Ω: that is 1/R, not R; 9 Ω: that is the series sum 6+3; 3 Ω: that is just one of the original values.
A 12 V cell with internal resistance 1 Ω is connected to a 5 Ω resistor. What current flows in the circuit?
Correct Answer: D. 2 A
• **2 A** = Total resistance is R + r = 5 + 1 = 6 Ω, so I = E/(R+r) = 12/6 = 2 A. • **I = 12/6 = 2 A** — Internal resistance reduces current compared to an ideal cell; terminal voltage = 12 − 2×1 = 10 V. • 💡 Wrong-option analysis: 12 A: ignores resistance entirely; 3 A: 12/4 uses wrong denominator; 1 A: only internal resistance in denominator.
A current of 3 A flows through a 4 Ω resistor. What power is dissipated in the resistor?
Correct Answer: A. 36 W
• **36 W** = Power P = I²R = 3² × 4 = 9 × 4 = 36 W. • **P = I²R = 9 × 4 = 36 W** — Doubling current quadruples power because of the I² factor. • 💡 Wrong-option analysis: 24 W: uses I×R×2, not I²R; 9 W: only I² without R factor; 12 W: I×R gives voltage not power.
A 10 V supply is connected across 2 Ω and 3 Ω in series. What is the potential difference across the 3 Ω resistor?
Correct Answer: A. 6 V
• **6 V** = Voltage divider: V₃ = 10 × 3/(2+3) = 10 × 3/5 = 6 V. • **V₃ = 10 × 3/5 = 6 V** — Series resistors share voltage in proportion to their resistance values. • 💡 Wrong-option analysis: 3 V: uses 10×3/10, wrong total; 4 V: uses wrong ratio; 5 V: equal split ignores different resistances.
Which statement is correct for Kirchhoff's current law at a junction?
Correct Answer: A. Sum of currents entering equals sum leaving
• **Sum of currents entering equals sum leaving** = KCL states that total current into a node equals total current leaving, based on charge conservation. • **ΣI_in = ΣI_out** — No charge accumulates at a steady-state junction; this allows analysis of branched circuits. • 💡 Wrong-option analysis: Resistance is same in all branches: false in general; Current is same in all branches: true only in series, not at a junction; Potential differences add to zero: that is KVL, not KCL.
Which statement is correct for Kirchhoff's voltage law in a closed loop?
Correct Answer: B. Algebraic sum of potential differences is zero
• **Algebraic sum of potential differences is zero** = KVL: the sum of all voltage rises and drops around any closed loop equals zero, from energy conservation. • **ΣV = 0 around any closed loop** — EMF sources add voltage; resistors drop voltage; together they balance to zero. • 💡 Wrong-option analysis: Resistance in loop is zero: not required; Sum of currents in loop is zero: that is KCL at a node; Charge in loop is zero: unrelated concept.
A Wheatstone bridge is balanced when which condition is satisfied?
Correct Answer: A. P/Q = R/S
• **P/Q = R/S** = At balance the galvanometer reads zero, implying the ratio P/Q equals R/S. • **P/Q = R/S (cross-multiply: PS = QR)** — This null condition allows precise resistance measurement without calibrating the galvanometer. • 💡 Wrong-option analysis: P/Q = S/R: ratio is inverted on one side; P − Q = R − S: difference rule is incorrect; P + Q = R + S: sum rule is incorrect.
An ideal ammeter should be connected how, and should have what resistance?
Correct Answer: D. In series and very low resistance
• **In series and very low resistance** = An ammeter must carry the full circuit current, so it is in series; very low resistance avoids altering the circuit. • **R_ideal ammeter → 0 Ω** — High resistance in series would reduce current drastically and give a wrong reading. • 💡 Wrong-option analysis: In parallel and very high resistance: voltmeter connection, not ammeter; In parallel and very low resistance: would short the branch; In series and very high resistance: would block current severely.
An ideal voltmeter should be connected how, and should have what resistance?
Correct Answer: D. In parallel and very high resistance
• **In parallel and very high resistance** = A voltmeter is connected across a component to measure voltage; very high resistance draws negligible current from the circuit. • **R_ideal voltmeter → ∞** — Low resistance in parallel would drain current and lower the measured voltage. • 💡 Wrong-option analysis: In parallel and very low resistance: would short the component; In series and very high resistance: blocks current, breaks the circuit; In series and very low resistance: would behave like a wire, not a meter.