JEE MAIN - Physics (2024 - 6th April Morning Shift - No. 29)
When a $$d c$$ voltage of $$100 \mathrm{~V}$$ is applied to an inductor, a $$d c$$ current of $$5 \mathrm{~A}$$ flows through it. When an ac voltage of $$200 \mathrm{~V}$$ peak value is connected to inductor, its inductive reactance is found to be $$20 \sqrt{3} \Omega$$. The power dissipated in the circuit is _________ W.
Answer
250
Explanation
To determine the power dissipated in the circuit, follow these steps:
- Calculate the resistance (R) using the DC current:
$ R = \frac{100 \, \text{V}}{5 \, \text{A}} = 20 \, \Omega $
- Determine the impedance (Z) by considering both resistance (R) and inductive reactance ($X_L$):
$ Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + (20\sqrt{3})^2} = 40 \, \Omega $
- Compute the peak current ($I_0$) using the AC peak voltage ($V_0$):
$ I_0 = \frac{V_0}{Z} = \frac{200 \, \text{V}}{40 \, \Omega} = 5 \, \text{A} $
- Calculate the power (P) using the RMS values of voltage and current, and consider the phase angle ($\cos \phi$):
$ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi = \frac{V_0 \cdot I_0}{2} \times \frac{R}{Z} = \frac{200 \cdot 5}{2} \times \frac{20}{40} = 250 \, \text{W} $
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