When a DC voltage of 100 V is applied to an inductor, a DC current of 5 A flows through it. When an AC voltage of 200 V peak value is connected to the inductor, its inductive reactance is found to be \( 20\sqrt{3} \, \Omega \). The power dissipated in the circuit is _________ W.
Correct Answer: 250
Approach Solution - 1
For DC voltage:
\[ R = \frac{V}{I} = \frac{100}{5} = 20 \, \Omega \]
For AC voltage:
\[ X_L = 20\sqrt{3} \, \Omega \]
\[ Z = \sqrt{X_L^2 + R^2} = \sqrt{(20\sqrt{3})^2 + 20^2} = \sqrt{1200 + 400} = 40 \, \Omega \]
Power dissipated in the circuit:
\[ P = I_\text{rms}^2 R = \left( \frac{V_\text{rms}}{Z} \right)^2 \times R \]
\[ P = \left( \frac{200}{\sqrt{2} \times 40} \right)^2 \times 20 \]
\[ P = \left( \frac{200}{40\sqrt{2}} \right)^2 \times 20 = 250 \, \text{W} \]
Approach Solution -2
Let's solve the problem step-by-step to find the power dissipated in the circuit.
First, we have a DC voltage of 100 V causing a DC current of 5 A to flow through the inductor. Since the current flows, the resistance \( R \) of the inductor can be determined using Ohm's law \( V = IR \). Thus:
\( R = \frac{V}{I} = \frac{100\,\text{V}}{5\,\text{A}} = 20\,\Omega \).
Now, consider the AC voltage scenario. Given the AC peak voltage \( V_0 = 200\,\text{V} \), the inductive reactance \( X_L = 20\sqrt{3}\,\Omega \) is provided. The impedance \( Z \) of the inductor in the AC circuit is therefore:
\( Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + (20\sqrt{3})^2} \).
Calculating each term:
\( 20^2 = 400 \).
\((20\sqrt{3})^2 = 400 \times 3 = 1200 \).
Thus:
\( Z = \sqrt{400 + 1200} = \sqrt{1600} = 40\,\Omega \).
Next, calculate the RMS value of the AC voltage \( V_{\text{rms}} \):
\( V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2}\,\text{V} \).
The RMS current \( I_{\text{rms}} \) through the inductor can be calculated as:
\( I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{100\sqrt{2}}{40} = \frac{5\sqrt{2}}{2}\,\text{A} \).
The power dissipated in the circuit, primarily due to the resistor, is given by \( P = I_{\text{rms}}^2 \times R \):
\( P = \left(\frac{5\sqrt{2}}{2}\right)^2 \times 20 \).
Calculating further:
\( \left(\frac{5\sqrt{2}}{2}\right)^2 = \frac{25 \times 2}{4} = 12.5 \).
Therefore,
\( P = 12.5 \times 20 = 250\,\text{W} \).
Hence, the power dissipated in the circuit is 250 W, which is within the given range of 250,250 W.
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