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A current of 4A flows in a coil when connected to 12V DC source. If the same coil is connected to a 12V, 50 rad/s source, a current of 2.4 A flows in the circuit. Then

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$R = 4 Ω$

L = 4H

L = 0.08 H

$R = 3 Ω$

answer is B, D.

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Detailed Solution

$\begin{array}{l} R = \frac{V_{DC}}{I_{DC}} = \frac{12}{4} = 3 Ω \end{array}$

When AC is applied,

$Z = \frac{V_{AC}}{I_{AC}} = \frac{12}{2.4} = 5 Ω$

$\begin{array}{l} ∴ X_{L} = \sqrt{Z^{2} - R^{2}} = \sqrt{5^{2} - 3^{2}} = 4 Ω \\ ∴ L = \frac{X_{L}}{ω} = \frac{4}{50} = 0 .08 H \end{array}$

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