An AC source is connected to an inductance of 100 mH, a capacitance of 100 μF and a resistance of 120 Ω as shown in figure. The time in which the resistance having a thermal capacity 2 J/°C will get heated by 16°C is _____________ s.

Solution

Physics sir · 4 steps

Find the rms current through the series LCR, then the heat Iᵣₘₛ² R t must supply C Δ T.

1. Reactances. From V = 20√2 sin(100t), ω = 100 rad s⁻¹. So X_L = ω L = 100 × 0.1 = 10 Ω and X_C = 1/(ω C) = 1/(100 × 100 × 10⁻⁶) = 100 Ω.
2. Impedance and current. Z = √(120² + (100 - 10)²) = √(14400 + 8100) = 150 Ω. The rms voltage is 20√2/√2 = 20 V, so Iᵣₘₛ = 20/150 = 2/15 A.
3. Heat balance. Only the resistor dissipates power, P = Iᵣₘₛ² R = 4/225 × 120 = 32/15 W. The heat needed is thermal capacity times temperature rise, 2 × 16 = 32 J.

   t = (C Δ T)/(Iᵣₘₛ² R) = 32/(32/15) = 15 s

4. So. The resistor heats up by 16°C in 15 s.

The common slip: 7.5. Using the peak current I₀ = 20√2/150 in I² R doubles the power and gives 7.5 s, not 15.

The idea to keep. Average power in an AC circuit is Iᵣₘₛ² R, and only the resistor uses it. Always convert the peak value in V₀ sin ω t to rms before finding the heat.

Answer from the NTA final answer key. Solution by PrepWiser.