Question

When \(100\,V\) DC is applied across a solenoid, current is \(1\,A\). When \(100\,V\) AC is applied, current is \(0.5\,A\). Frequency \(=50\,Hz\). Find inductance \(= x\,mH\).

Hint:

Use DC to find resistance and AC to find impedance. Then \(X_L = \sqrt{Z^2 - R^2}\).

Answer 1

Correct Answer: 550

Concept:
•DC: Solenoid acts as pure resistor \(\Rightarrow R = \frac{V_{DC}}{I_{DC}}\)
•AC: Impedance \(Z = \frac{V_{AC}}{I_{AC}} = \sqrt{R^2 + X_L^2}\)
•\(X_L = \omega L = 2\pi f L\)

Step 1:Find resistance: \[ R = \frac{100}{1} = 100\,\Omega \]

Step 2:Find impedance: \[ Z = \frac{100}{0.5} = 200\,\Omega \]

Step 3:Using \(Z^2 = R^2 + X_L^2\): \[ 200^2 = 100^2 + X_L^2 \] \[ 40000 = 10000 + X_L^2 \] \[ X_L^2 = 30000 \] \[ X_L = \sqrt{30000} = 100\sqrt{3} \approx 173.2\,\Omega \]

Step 4:Find inductance: \[ X_L = \omega L = 2\pi f L \] \[ 173.2 = 2\pi \times 50 \times L \] \[ 173.2 = 100\pi \times L \] \[ L = \frac{173.2}{100\pi} \approx \frac{173.2}{314.16} \approx 0.551\,H \] \[ L \approx 550\,mH \] \[ \Rightarrow x = 550 \]