Question

When 100 V dc is applied across a solenoid, a current of 1 A flows in it. When 100 V ac is applied across the same solenoid the current drops to 0.5 A. If the frequency of the ac source is 50 Hz, the impedance and inductance of the solenoid are:

• A. 200 \(\Omega\) and 0.55 H
• B. 100 \(\Omega\) and 0.86 H
• C. 200 \(\Omega\) and 1.0 H
• D. 100 \(\Omega\) and 0.93 H

Hint:

For a solenoid in an AC circuit, the impedance is the sum of the resistance and the inductive reactance, and the inductance can be calculated from the reactance.

Answer 1

The Correct Option is A

For the DC case, the impedance is simply the resistance \(R\), and the current is given by: \[ I_{\text{dc}} = \frac{V}{R} \quad \Rightarrow \quad R = \frac{V}{I_{\text{dc}}} = \frac{100}{1} = 100 \, \Omega \] For the AC case, the total impedance \(Z\) is the sum of the resistance \(R\) and the inductive reactance \(X_L\): \[ I_{\text{ac}} = \frac{V}{Z} \quad \Rightarrow \quad Z = \frac{V}{I_{\text{ac}}} = \frac{100}{0.5} = 200 \, \Omega \] The impedance \(Z\) for an inductive circuit is given by: \[ Z = \sqrt{R^2 + X_L^2} \] Substitute the values: \[ 200 = \sqrt{100^2 + X_L^2} \quad \Rightarrow \quad X_L = \sqrt{200^2 - 100^2} = 173.2 \, \Omega \] The inductive reactance \(X_L\) is related to the inductance \(L\) by: \[ X_L = 2 \pi f L \] Substitute the frequency \(f = 50 \, \text{Hz}\): \[ 173.2 = 2 \pi \times 50 \times L \quad \Rightarrow \quad L = \frac{173.2}{2 \pi \times 50} \approx 0.55 \, \text{H} \]