Question

When 100 V DC source is applied across a solenoid, a steady current of 1A flows in it. When 100 V AC source is applied across the same solenoid, current drops to 0.5A. If the frequency of AC source is $\frac{150\sqrt{3}}{\pi}$ Hz. Find inductance of the solenoid.

Answer 1

1/3 H

Answer 2

The problem involves analyzing the behavior of a solenoid under both DC and AC voltage sources to determine its inductance.

1. Determine Resistance (R) using DC Source:

When a DC source is applied, the frequency (f) is zero. Therefore, the inductive reactance ($X_L = 2\pi f L$) is zero. The solenoid behaves purely as a resistor.

Given:

DC Voltage ($V_{DC}$) = 100 V DC Current ($I_{DC}$) = 1 A

Using Ohm's Law:

$R = \frac{V_{DC}}{I_{DC}}$

$R = \frac{100 \, \text{V}}{1 \, \text{A}}$

$R = 100 \, \Omega$

2. Determine Impedance (Z) using AC Source:

When an AC source is applied, the solenoid acts as an R-L series circuit, where R is the resistance and L is the inductance.

Given:

AC Voltage ($V_{AC}$) = 100 V AC Current ($I_{AC}$) = 0.5 A

Using Ohm's Law for AC circuits (V = I * Z):

$Z = \frac{V_{AC}}{I_{AC}}$

$Z = \frac{100 \, \text{V}}{0.5 \, \text{A}}$

$Z = 200 \, \Omega$

3. Determine Inductive Reactance ($X_L$):

For an R-L series circuit, the impedance (Z) is related to resistance (R) and inductive reactance ($X_L$) by the formula:

$Z = \sqrt{R^2 + X_L^2}$

Squaring both sides:

$Z^2 = R^2 + X_L^2$

Rearranging to find $X_L$:

$X_L^2 = Z^2 - R^2$

$X_L = \sqrt{Z^2 - R^2}$

Substitute the calculated values of Z and R:

$X_L = \sqrt{(200 \, \Omega)^2 - (100 \, \Omega)^2}$

$X_L = \sqrt{40000 - 10000}$

$X_L = \sqrt{30000}$

$X_L = \sqrt{3 \times 10000}$

$X_L = 100\sqrt{3} \, \Omega$

4. Determine Inductance (L):

The inductive reactance ($X_L$) is also given by the formula:

$X_L = 2\pi f L$

Given frequency (f) = $\frac{150\sqrt{3}}{\pi}$ Hz

Rearranging to find L:

$L = \frac{X_L}{2\pi f}$

Substitute the values of $X_L$ and f:

$L = \frac{100\sqrt{3} \, \Omega}{2\pi \left(\frac{150\sqrt{3}}{\pi} \, \text{Hz}\right)}$

$L = \frac{100\sqrt{3}}{2 \times 150\sqrt{3}}$

$L = \frac{100\sqrt{3}}{300\sqrt{3}}$

$L = \frac{100}{300}$

$L = \frac{1}{3} \, \text{H}$

The inductance of the solenoid is $\frac{1}{3}$ H.