JEE MAIN - Physics (2025 - 7th April Evening Shift - No. 23)
Explanation
To calculate the average power dissipated in the circuit, we follow these steps:
Formula for Power:
$ P = V_{\text{rms}} \times I_{\text{rms}} \times \cos \phi $
Expressing Power in Terms of Voltage and Impedance:
$ P = V_{\text{rms}} \times \frac{V_{\text{rms}}}{Z} \times \frac{R}{Z} $
Simplifying the Expression:
$ P = \frac{V_{\text{rms}}^2 \times R}{Z^2} $
Calculate Impedance (Z):
The impedance of the circuit (Z) is calculated as:
$ Z = \sqrt{R^2 + (X_L - X_C)^2} $
Given:
- $ X_L = 100 \, \Omega $
- $ X_C = 50 \, \Omega $
- $ R = 50 \, \Omega $
$ X_L - X_C = 100 \, \Omega - 50 \, \Omega = 50 \, \Omega $
Thus:
$ Z = \sqrt{50^2 + 50^2} = 50\sqrt{2} \, \Omega $
Substitute Values to Find Power (P):
Given $ V_{\text{rms}} = 10 \, \text{V} $, we calculate $ P $ as follows:
$ P = \frac{10^2 \times 50}{(50\sqrt{2})^2} $
Simplifying further:
$ P = \frac{100 \times 50}{2500 \times 2} = 1 \, \text{W} $
Therefore, the average power dissipated by the circuit is $ 1 \, \text{W} $.
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