Question

Given circuit contain 3 resistors, 2 capacitors and 2 inductors. Value of reactances of capacitors and inductors also shown in the figure. Circuit is connected to a 20 V AC source. Find rate of heat dissipation (in watts) in the circuit.

• A. 13.23
• B. 10.50
• C. 25.75
• D. 30.00

Answer 1

Answer: (A)

13.23

Answer 2

The rate of heat dissipation in an AC circuit occurs only in resistors. The total heat dissipation is the sum of the power dissipated by each resistor.

The circuit consists of three parallel branches connected in series with a $10 \Omega$ resistor and the AC source (which has an additional $20 \Omega$ series resistance). The AC source voltage is $V_{rms} = 20$ V.

The three parallel branches have the following impedances:

1. Branch 1: $Z_1 = jX_{L1} - jX_{C1} = j4 - j3 = j1 \Omega$. (Purely reactive, no power dissipation)
2. Branch 2: $Z_2 = R_2 + jX_{L2} = 6 + j8 \Omega$. The magnitude is $|Z_2| = \sqrt{6^2 + 8^2} = 10 \Omega$.
3. Branch 3: $Z_3 = R_3 - jX_{C2} = 3 - j4 \Omega$. The magnitude is $|Z_3| = \sqrt{3^2 + (-4)^2} = 5 \Omega$.

The total series resistance in the circuit is $R_{series} = 10 \Omega + 20 \Omega = 30 \Omega$.

The equivalent impedance of the parallel combination ($Z_{parallel}$) is calculated as: $\frac{1}{Z_{parallel}} = \frac{1}{Z_1} + \frac{1}{Z_2} + \frac{1}{Z_3} = \frac{1}{j1} + \frac{1}{6+j8} + \frac{1}{3-j4}$ $\frac{1}{Z_{parallel}} = -j + \frac{6-j8}{100} + \frac{3+j4}{25} = -j + (0.06 - j0.08) + (0.12 + j0.16)$ $\frac{1}{Z_{parallel}} = 0.18 - j0.92$ $Z_{parallel} = \frac{1}{0.18 - j0.92} \approx 0.2048 + j1.047 \Omega$.

The total impedance of the circuit is $Z_{total} = R_{series} + Z_{parallel} = 30 + (0.2048 + j1.047) = 30.2048 + j1.047 \Omega$. The magnitude of the total impedance is $|Z_{total}| = \sqrt{30.2048^2 + 1.047^2} \approx 30.223 \Omega$.

The RMS current from the source is $I_{rms} = \frac{V_{rms}}{|Z_{total}|} = \frac{20}{30.223} \approx 0.6617$ A.

Power dissipated by the series resistors ($10 \Omega$ and $20 \Omega$): $P_{series} = I_{rms}^2 \times R_{series} = (0.6617)^2 \times 30 \approx 13.1355$ W.

The voltage across the parallel combination is $V_{parallel} = I_{rms} \times Z_{parallel}$. $|V_{parallel}| = |I_{rms} \times Z_{parallel}| \approx 0.6617 \times |0.2048 + j1.047| \approx 0.6617 \times \sqrt{0.2048^2 + 1.047^2} \approx 0.6617 \times 1.065 \approx 0.705$ V.

Power dissipated in Branch 2 resistor ($R_2 = 6 \Omega$): $|I_2| = \frac{|V_{parallel}|}{|Z_2|} = \frac{0.705}{10} = 0.0705$ A. $P_{R2} = |I_2|^2 R_2 = (0.0705)^2 \times 6 \approx 0.00497 \times 6 \approx 0.0298$ W.

Power dissipated in Branch 3 resistor ($R_3 = 3 \Omega$): $|I_3| = \frac{|V_{parallel}|}{|Z_3|} = \frac{0.705}{5} = 0.141$ A. $P_{R3} = |I_3|^2 R_3 = (0.141)^2 \times 3 \approx 0.0199 \times 3 \approx 0.0597$ W.

Total heat dissipation $P_{total} = P_{series} + P_{R2} + P_{R3} \approx 13.1355 + 0.0298 + 0.0597 \approx 13.225$ W. Rounding to two decimal places, the heat dissipation is approximately 13.23 W.