Electricity — Set 3

Correct Answer: A. Diode

• **Diode** = A diode is a p-n junction device that conducts in the forward direction and blocks reverse current. • **Forward biased: low resistance; reverse biased: very high resistance** — Used in rectifiers to convert AC to DC. • 💡 Wrong-option analysis: Inductor: stores energy in a magnetic field; Capacitor: stores charge and blocks DC; Transformer: changes AC voltage levels via mutual induction.

Correct Answer: D. 2 Ω

• **2 Ω** = Using Req = R1R2/(R1+R2) = 3×6/(3+6) = 18/9 = 2 Ω. • **Req = 18/9 = 2 Ω** — Parallel equivalent is always less than the smallest resistor (3 Ω here). • 💡 Wrong-option analysis: 4.5 Ω: arithmetic mean (3+6)/2, not the parallel formula; 9 Ω: series sum 3+6; 1 Ω: no valid arithmetic gives this.

Correct Answer: B. Sum of currents entering equals sum leaving

• **Sum of currents entering equals sum leaving** = KCL: the algebraic sum of currents at any junction is zero, reflecting conservation of charge. • **ΣI_in = ΣI_out** — KCL is used to write node equations in complex circuits. • 💡 Wrong-option analysis: Voltage same in all branches: property of parallel connections, not KCL; Resistance is maximum: resistance has no specific value at a junction; Power conserved in each branch: general energy principle, not the statement of KCL.

Correct Answer: D. Farad

• **Farad** = Capacitance C = Q/V is measured in farad; 1 F = 1 C/V. • **1 F = 1 C/V** — Practical capacitors range from picofarads (pF) to millifarads (mF). • 💡 Wrong-option analysis: Ohm: unit of resistance; Henry: unit of inductance; Tesla: unit of magnetic flux density.

Correct Answer: A. 4 μF

• **4 μF** = Using C = Q/V = 12/3 = 4 μF. • **C = Q/V = 12 μC/3 V = 4 μF** — Doubling the voltage on the same capacitor doubles the stored charge. • 💡 Wrong-option analysis: 0.25 μF: inverts to V/Q; 9 μF: subtracts 12−3; 36 μF: multiplies Q×V instead of dividing.

Correct Answer: C. Increases capacitance

• **Increases capacitance** = A dielectric with relative permittivity κ > 1 reduces the effective field and raises capacitance to C = κC₀. • **C = κε₀A/d; κ > 1 always increases C** — Dielectrics also raise the maximum safe voltage. • 💡 Wrong-option analysis: Reverses polarity: a dielectric cannot reverse stored charge polarity; Decreases capacitance: always increases, never decreases; Makes capacitance zero: a conducting plate would short the capacitor, not a dielectric.

Correct Answer: B. 24 W

• **24 W** = Using P = VI = 12 × 2 = 24 W. • **P = VI = 12 × 2 = 24 W** — Same result from P = I²R using R = V/I = 6 Ω. • 💡 Wrong-option analysis: 144 W: uses P = V² without dividing by R; 6 W: V/I gives resistance not power; 10 W: no valid formula gives this.

Correct Answer: B. Vrms = V0/√2

• **Vrms = V0/√2** = For a sine wave, the RMS voltage is peak voltage divided by √2. • **Vrms = V0/√2 ≈ 0.707 V0** — India's 230 V RMS supply means peak voltage ≈ 325 V. • 💡 Wrong-option analysis: V0/2: average of a half-wave, not the RMS; √2 V0: peak from RMS, which is the inverse relationship; V0: true only for DC, not a sinusoid.

Correct Answer: B. Provide a low resistance path to ground for safety

• **Provide a low resistance path to ground for safety** = The earth wire connects the metal body of an appliance to ground so leakage current flows safely to earth instead of through a person. • **Earth wire → body held at ~0 V; trips fuse/breaker on fault** — Without earthing, a fault can make the appliance body live at mains voltage. • 💡 Wrong-option analysis: Make appliances consume less power: earthing does not reduce normal power consumption; Decrease current in all cases: earthing does not reduce operating current; Increase voltage: earthing has no connection to supply voltage.

Correct Answer: C. 10 Ω

• **10 Ω** = In series, Req = 2 + 3 + 5 = 10 Ω. • **Req = R1 + R2 + R3 = 10 Ω** — Series combination always gives resistance larger than any single resistor. • 💡 Wrong-option analysis: 6 Ω: adds only 2+3, forgets the 5; 30 Ω: product 2×3×5, not the sum; 1/10 Ω: result of parallel formula.