Circuits — Set 5

Correct Answer: D. G = 1/R

• **G = 1/R** = Conductance is defined as the reciprocal of resistance; higher resistance means lower conductance. • **G = 1/R in siemens (S)** — A 10 Ω resistor has conductance 0.1 S; a 0.1 Ω conductor has conductance 10 S. • 💡 Wrong-option analysis: G = R: same value as resistance, but conductance and resistance are reciprocals; G = R²: no standard formula; G = 1/R²: reciprocal of R², not R.

Correct Answer: B. Work done by the source per unit charge

• **Work done by the source per unit charge** = EMF (ε) = W/Q; the source converts chemical, mechanical, or other energy into electrical energy per coulomb. • **ε = W/Q in volts (V = J/C)** — EMF is an intrinsic source property; terminal voltage is what's available after internal resistance drop. • 💡 Wrong-option analysis: Charge per unit work: inverted definition, gives 1/V; Power per unit charge: power/charge = (W/t)/Q = W/(Qt) = V/t, not emf; Current per unit charge: A/C = 1/s, a frequency, not emf.

Correct Answer: A. Its emf

• **Its emf** = When I = 0, internal drop Ir = 0, so terminal voltage V = E − Ir = E. • **Open circuit: V_terminal = E (emf)** — This is how emf is measured: using a high-resistance voltmeter that draws negligible current. • 💡 Wrong-option analysis: Its power rating: power is not a voltage; Zero: terminal voltage equals emf when no current flows; Its internal drop Ir: Ir = 0 when I = 0, so this option is itself zero.

Correct Answer: D. P/Q = R/S

• **P/Q = R/S** = Balance condition: no current through galvanometer, so junction potentials are equal, giving P/Q = R/S (or equivalently PS = QR). • **Balance: P/Q = R/S → unknown = Q×R/P** — Null method avoids galvanometer calibration; very precise resistance measurements possible. • 💡 Wrong-option analysis: P − Q = R − S: difference rule is not the bridge balance condition; P + Q = R + S: sum rule does not ensure zero galvanometer current; P = Q = R = S: balance occurs for many other ratio combinations, not only when all four are equal.

Correct Answer: B. Low melting point

• **Low melting point** = Fuse wire must melt quickly on excess current; a low melting point alloy (tin-lead) ensures rapid circuit interruption. • **Common fuse alloy: tin-lead, melting point ~183–300 °C** — High resistivity also helps generate heat quickly at the rated excess current. • 💡 Wrong-option analysis: Very high resistivity and very high melting point: high melting point would resist melting and defeat the purpose; Very high melting point: would not melt at overcurrent levels, offering no protection; Perfect superconductivity: superconductors have zero resistance and would not heat up at all.

Correct Answer: B. To protect the circuit by breaking it during excessive current

• **To protect the circuit by breaking it during excessive current** = A fuse melts when current exceeds its rating, opening the circuit and preventing wiring or device damage. • **Fuse: sacrificial overcurrent protection device** — Always placed in series on the live wire; must be replaced after it operates. • 💡 Wrong-option analysis: To increase current: fuses do not amplify current; To store electrical energy: that is a capacitor or battery; To measure voltage: that is a voltmeter.

Correct Answer: A. An open circuit

• **An open circuit** = In steady DC, a fully charged capacitor has no current flow; it behaves exactly like an open circuit (infinite impedance at f = 0). • **X_C = 1/(2πfC); at f = 0 (DC), X_C → ∞** — Capacitors block DC but pass AC; this is why they are used as coupling/decoupling elements. • 💡 Wrong-option analysis: A constant current source: current drops to zero in steady DC; A short circuit: that would apply at very high frequencies, not DC; A resistor of fixed value: capacitor impedance depends on frequency, not a fixed resistance.

Correct Answer: D. τ = RC

• **τ = RC** = The time constant of an RC circuit is the product of resistance in ohms and capacitance in farads, giving τ in seconds. • **τ = RC seconds; larger τ → slower charging/discharging** — At t = τ, capacitor charges to 63% (or discharges to 37%) of its value. • 💡 Wrong-option analysis: τ = R + C: adding different units (Ω + F) is dimensionally nonsensical; τ = R/C: gives Ω/F = s⁻¹, not seconds; τ = C/R: gives F/Ω = s⁻¹, not seconds.

Correct Answer: B. Charge on each is the same

• **Charge on each is the same** = In series, only one current path exists; the same charge Q flows onto each capacitor plate. • **Q_series = same for all; voltages differ: V = Q/C** — Smaller capacitor carries the larger voltage share in series. • 💡 Wrong-option analysis: Voltage across each is the same: equal voltage applies to parallel capacitors, not series; Capacitance of each becomes same: series connection does not change individual capacitance values; Energy stored in each is always same: U = Q²/(2C) differs if capacitances differ.

Correct Answer: D. It is less than the smallest branch resistance

• **It is less than the smallest branch resistance** = Adding parallel paths always increases current flow; mathematically, 1/R_eq = sum of 1/Rᵢ > 1/R_min, so R_eq < R_min. • **R_eq < R_smallest (strictly)** — Even adding a very large resistance in parallel slightly reduces R_eq. • 💡 Wrong-option analysis: It becomes zero for any values: zero only with a short-circuit (R = 0) branch; It equals the sum of resistors: sum rule is for series; It equals the largest resistor: parallel always reduces below the smallest, not equals the largest.