Circuits — Set 2
Physics · परिपथ · Questions 11–20 of 50
Heat produced in a resistor R by a current I in time t is given by which expression?
Correct Answer: A. H = I^2Rt
• **H = I²Rt** = Joule's law of heating: energy dissipated equals current squared times resistance times time. • **H = I²Rt (in joules)** — Note that H = VIt is also correct since V = IR, but from the given options only H = I²Rt is in the exact listed form. • 💡 Wrong-option analysis: H = I²R/t: divides by time instead of multiplying; H = IRt: missing the square on I; H = VIt: equivalent but not in the form listed as option A here.
What does the current rating of a fuse wire represent?
Correct Answer: A. Maximum safe current before it melts
• **Maximum safe current before it melts** = A fuse rating specifies the highest current it can carry without melting and breaking the circuit. • **Fuse melts when I > rated value** — Made of low-melting-point alloy so excess current generates enough heat to fuse the wire instantly. • 💡 Wrong-option analysis: Current at which voltage becomes zero: unrelated; Minimum current for melting: rating is the maximum safe, not the minimum to melt; Average current in circuit: fuse rating is not about average current.
Two identical bulbs are connected in series to a battery. Compared to a single bulb, each bulb glows how?
Correct Answer: A. Dimmer than the single bulb
• **Dimmer than the single bulb** = In series, total resistance doubles, current halves, and power per bulb becomes P = I²R = (I/2)²R = one-quarter of the single-bulb power. • **P ∝ I², current halves → power = 1/4 of original** — Both bulbs share the voltage, so each gets only half. • 💡 Wrong-option analysis: Brighter than the single bulb: impossible with same battery and higher resistance; With the same brightness: same only if voltage per bulb were the same; Only one bulb glows: both glow, just dimmer.
If the length of a uniform wire is doubled while keeping its area and material the same, what happens to its resistance?
Correct Answer: D. It becomes twice
• **It becomes twice** = Resistance R = ρL/A; doubling L with constant ρ and A gives 2R. • **R = ρL/A → 2R when L → 2L** — Resistivity ρ depends only on material and temperature, not on dimensions. • 💡 Wrong-option analysis: It becomes zero: length increase can never make resistance zero; It becomes four times: that would require both L doubled and A halved; It becomes half: halving happens if length is halved.
What is the SI unit of electrical resistivity?
Correct Answer: C. Ω·m
• **Ω·m** = From R = ρL/A, rearranging gives ρ = RA/L; units are (Ω × m²)/m = Ω·m. • **ρ unit = Ω·m** — Resistivity is a material property; copper ≈ 1.7×10⁻⁸ Ω·m, glass ≈ 10¹² Ω·m. • 💡 Wrong-option analysis: Ω: that is resistance, not resistivity; Ω/s: no time dimension in resistivity; Ω/m: would apply if resistivity were R/L, but it is RA/L.
For most metallic conductors, how does resistance change when temperature increases?
Correct Answer: C. Resistance increases
• **Resistance increases** = In metals, higher temperature means more lattice vibrations, more electron scattering, and higher resistivity. • **R = R₀(1 + αΔT), α > 0 for metals** — This positive temperature coefficient makes metals useful as temperature sensors (RTDs). • 💡 Wrong-option analysis: Resistance decreases to zero: that describes superconductors below critical temperature; Resistance stays exactly constant: true only for specific alloys like Nichrome over small ranges; Resistance becomes unpredictable: metals follow a well-defined linear relationship.
In an RC charging circuit, the time constant τ corresponds to what change in capacitor voltage?
Correct Answer: C. It reaches about 63% of its final value
• **It reaches about 63% of its final value** = At t = τ = RC, V_C = V₀(1 − e⁻¹) ≈ V₀ × 0.632. • **τ = RC; at t = τ, V_C ≈ 63% of final** — After 5τ, the capacitor is considered fully charged (>99%). • 💡 Wrong-option analysis: It reaches 50% of its final value: that occurs at t = 0.693τ, the half-life, not τ; It becomes zero: voltage is zero only at t = 0; It becomes maximum instantly: charging is exponential, never instantaneous.
Two capacitors of 2 μF and 3 μF are connected in series. What is the equivalent capacitance?
Correct Answer: D. 6/5 μF
• **6/5 μF** = For series capacitors, 1/C_eq = 1/2 + 1/3 = 3/6 + 2/6 = 5/6, so C_eq = 6/5 μF. • **C_eq = 6/5 μF = 1.2 μF** — Series connection always reduces total capacitance below the smallest value. • 💡 Wrong-option analysis: 1 μF: incorrect arithmetic; 5 μF: that is the parallel sum 2+3; 1/5 μF: that is 1/C_eq, not C_eq.
Two capacitors of 2 μF and 3 μF are connected in parallel. What is the equivalent capacitance?
Correct Answer: A. 5 μF
• **5 μF** = For parallel capacitors, capacitances add: C_eq = 2 + 3 = 5 μF. • **C_eq = 2 + 3 = 5 μF** — Each capacitor gets the same voltage; together they store more charge, increasing total capacitance. • 💡 Wrong-option analysis: 1 μF: no standard formula gives this; 6 μF: product 2×3 is not the parallel rule; 6/5 μF: that is the series result, not parallel.
Which formula gives the energy stored in a capacitor of capacitance C at voltage V?
Correct Answer: D. U = (1/2)CV^2
• **U = (1/2)CV²** = Energy stored in a capacitor equals half the capacitance times voltage squared. • **U = (1/2)CV² = Q²/(2C)** — The factor 1/2 arises because voltage builds gradually from 0 to V during charging. • 💡 Wrong-option analysis: U = Q/C: that equals voltage V, not energy; U = QV: missing the factor 1/2; U = CV: that gives charge Q = CV, not energy.