Question

A coil has a resistance of \( 30 \, \Omega \) and an inductive reactance of \( 20 \, \Omega \) at 50 Hz frequency. If an AC source of 200 V and 100 Hz is connected across the coil, then how much current will flow through the coil?

• A. \( 3 \, \mathrm{A} \)
• B. \( 4 \, \mathrm{A} \)
• C. \( 5 \, \mathrm{A} \)
• D. \( 6 \, \mathrm{A} \)

Hint:

For AC circuits with resistance and reactance, use the formula \( Z = \sqrt{R^2 + X_L^2} \) to calculate impedance. Then apply Ohm's Law, \( I = \frac{V}{Z} \), to find the current.

Answer 1

The Correct Option is A

Given the problem statement, the inductive reactance at \( 50 \, \mathrm{Hz} \) is \( X_L = 20 \, \Omega \). The formula for inductive reactance is \( X_L = 2\pi f L \). Substituting the known values, \( 2\pi \times 50 \times L = 20 \). Solving for inductance \( L \), we get \( L = \frac{20}{2\pi \times 50} = \frac{1}{5\pi} \, \mathrm{H} \). When the frequency is increased to \( 100 \, \mathrm{Hz} \), the new inductive reactance is calculated using \( X_L = 2\pi f L \). Substituting \( f = 100 \, \mathrm{Hz} \) and \( L = \frac{1}{5\pi} \, \mathrm{H} \), we find \( X_L = 2\pi \times 100 \times \frac{1}{5\pi} = 40 \, \Omega \). The circuit impedance \( Z \) is determined by the formula \( Z = \sqrt{R^2 + X_L^2} \). With \( R = 30 \, \Omega \) and \( X_L = 40 \, \Omega \), the impedance is \( Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \Omega \). Ohm's Law, \( I = \frac{V}{Z} \), is used to find the current. Substituting \( V = 200 \, \mathrm{V} \) and \( Z = 50 \, \Omega \), the current is \( I = \frac{200}{50} = 4 \, \mathrm{A} \). Therefore, the current flowing through the coil is \( I = 4 \, \mathrm{A} \).