Question

An inductor of reactance 10 and a resistor of resistance 3 are connected in series to the terminals of 10 V (rms) AC source. The power dissipated in the circuit is-

Answer 1

$\frac{300}{109}$ W (approximately 2.75 W)

Answer 2

Solution:

1. Calculate the Total Impedance (Z):

   $$Z = \sqrt{R^2 + X_L^2} = \sqrt{3^2 + 10^2} = \sqrt{9+100} = \sqrt{109}$$

2. Determine the Circuit Current (I):

   $$I = \frac{V}{Z} = \frac{10}{\sqrt{109}}$$

3. Find the Power Dissipated (P):

   Only the resistor dissipates real power. The power is given by:

   $$P = I^2R = \left(\frac{10}{\sqrt{109}}\right)^2 \times 3 = \frac{100 \times 3}{109} = \frac{300}{109}\, \text{W}$$

   Thus, the power dissipated in the circuit is $\frac{300}{109}$ watts (approximately $2.75$ W).

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Core Explanation:

• Impedance: $Z = \sqrt{R^2 + X_L^2}$.
• Current: $I = \frac{V}{Z}$.
• Power: $P = I^2R = \frac{V^2R}{Z^2}$.