Circuits — Set 3

Correct Answer: C. Energy

• **Energy** = 1 kWh = power of 1000 W used for 3600 s = 3,600,000 J; it is a unit of energy. • **1 kWh = 3.6 × 10⁶ J** — Electricity bills use kWh because joules are too small for practical household use. • 💡 Wrong-option analysis: Electric charge: measured in coulombs; Power: measured in watts or kilowatts, not watt-hours; Electric current: measured in amperes.

Correct Answer: C. To provide a low-resistance path for leakage current

• **To provide a low-resistance path for leakage current** = Earthing ensures any leakage to the metal body flows safely to ground, keeping the body near zero potential. • **Earth wire → fault current → ground, trips fuse/breaker** — Without earthing, a fault makes the metal body live and anyone touching it gets a shock. • 💡 Wrong-option analysis: To increase voltage: earthing does not affect supply voltage; To stop all current in the circuit: it provides a path, does not stop current; To increase resistance: earth wire is a low-resistance conductor.

Correct Answer: B. It allows current mainly in one direction

• **It allows current mainly in one direction** = A p-n junction diode conducts in forward bias and blocks in reverse bias, acting as a one-way valve for current. • **Forward biased: low resistance; reverse biased: very high resistance** — This property is used for rectification of AC to DC. • 💡 Wrong-option analysis: It converts heat into electricity: that is a thermoelectric device; It stores charge: that is a capacitor; It increases resistance: a diode does not simply increase resistance.

Correct Answer: C. It is the same through all resistors

• **It is the same through all resistors** = In a series circuit there is only one path; the same charge must flow through every element. • **I_series = constant throughout** — KCL at each node confirms no current can split when there is a single loop. • 💡 Wrong-option analysis: It depends only on one resistor: total resistance determines current, not just one; It splits equally: current splits only in parallel branches; It is zero in some resistors: if any resistor were open, the entire circuit would have zero current.

Correct Answer: B. It is the same across each branch

• **It is the same across each branch** = All parallel branches share the same two nodes, so voltage across each is equal regardless of resistance values. • **V_parallel = common voltage across all branches** — This is why household appliances in parallel all receive 220 V. • 💡 Wrong-option analysis: It is different only for identical resistors: same voltage for all parallel branches, not just identical ones; It depends on current only: voltage equals current times resistance, not current alone; It is zero across each branch: zero voltage would mean no current and no functioning branches.

Correct Answer: C. V = E - Ir

• **V = E − Ir** = Terminal voltage equals emf minus the internal voltage drop Ir; internal resistance always reduces available output voltage. • **V = E − Ir (drops with increasing current)** — When I = 0 (open circuit), V = E; under load V < E. • 💡 Wrong-option analysis: V = E/r: divides emf by internal resistance, dimensionally wrong for voltage; V = Ir − E: negative of correct expression, gives negative value for normal operation; V = E + Ir: would mean voltage rises with current, opposite of reality.

Correct Answer: B. R = r

• **R = r** = Power delivered to load P = E²R/(R+r)²; differentiating with respect to R and setting to zero gives R = r. • **P_max when R = r; P_max = E²/(4r)** — At this condition efficiency is 50%; maximum efficiency (not maximum power) requires R >> r. • 💡 Wrong-option analysis: R is much larger than r: maximises efficiency, not power delivered; R = 0: short circuit gives zero voltage across load, zero power to load; r is much smaller than R: same as option A, gives high efficiency but not maximum power to load.

Correct Answer: C. Wheatstone bridge principle

• **Wheatstone bridge principle** = A meter bridge is a practical implementation of the Wheatstone bridge using a uniform resistance wire instead of discrete resistors. • **Balance condition: R_unknown/R_known = l/(100−l)** — Null-deflection method avoids galvanometer calibration errors. • 💡 Wrong-option analysis: Electrostatic induction: unrelated to resistance measurement; Potentiometer principle: potentiometers measure voltage, not resistance directly; Ohm's law only: Ohm's law is used within the derivation, but the core principle is the bridge balance.

Correct Answer: B. R/2

• **R/2** = For two equal resistors in parallel, 1/R_eq = 1/R + 1/R = 2/R, so R_eq = R/2. • **R_eq = R/2** — Parallel combination always gives half the resistance when both resistors are equal. • 💡 Wrong-option analysis: R²: product of resistances has wrong units; 2R: that is the series combination; R: same as each individual resistor, but parallel always reduces total resistance.

Correct Answer: B. It is nearly zero

• **It is nearly zero** = An ideal ammeter has zero resistance; with V = IR and R → 0, the voltage drop is negligible. • **V_ammeter = I × R_ammeter ≈ 0** — Any significant voltage drop would reduce circuit current and give incorrect reading. • 💡 Wrong-option analysis: It is very large: high voltage drop means high resistance, opposite of ideal ammeter; It is equal to emf: that would mean all voltage drops across ammeter, leaving nothing for load; It is always negative: voltage sign depends on current direction, not a defining property.