A 750 Hz, 20 V (rms) source is connected to a resistance of 100 Ω, an inductance of 0.1803 H and a capacitance of 10 μF all in series. The time in which the resistance (heat capacity 2 J/ᵒC) will get heated by 10ᵒC. (assume no loss of heat to the surroudnings) is close to :

Solution

Physics sir · 4 steps

Only the resistor heats up, so find the current through it and use Iᵣₘₛ² R as the heating rate.

1. Find the reactances. ω = 2π × 750 = 4712 rad s⁻¹. So X_L = ω L = 4712 × 0.1803 = 849.6 Ω and X_C = 1/(ω C) = 1/(4712 × 10⁻⁵) = 21.2 Ω.
2. Impedance. X_L - X_C = 828.4 Ω. So Z² = 100² + 828.4² = 6.963 × 10⁵ Ω², which gives Z ≈ 834 Ω.

   P = Iᵣₘₛ² R = (Vᵣₘₛ² R)/Z² = (20² × 100)/(6.963 × 10⁵) ≈ 0.0574 W

3. Heat needed. The heat needed is heat capacity times the temperature rise, Q = 2 × 10 = 20 J. The inductor and capacitor take no average power, so all of P goes into the resistor.
4. So. t = Q/P = 20/0.0574 ≈ 348 s, which is option (a).

The idea to keep. In an LCR circuit only R takes average power. Find Z with both reactances, then use P = Vᵣₘₛ² R / Z². Heat capacity times temperature rise gives the energy needed.

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