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

According to the problem, the inductive reactance at \( 50 \, \mathrm{Hz} \) is: \[ X_L = 20 \, \Omega. \] The inductive reactance \( X_L \) is given by: \[ X_L = 2\pi f L. \] Substitute \( f = 50 \, \mathrm{Hz} \) and \( X_L = 20 \, \Omega \): \[ 2\pi \times 50 \times L = 20. \] Simplify: \[ L = \frac{20}{2\pi \times 50} = \frac{1}{5\pi} \, \mathrm{H}. \] If the frequency is increased to \( 100 \, \mathrm{Hz} \), the new inductive reactance is: \[ X_L = 2\pi f L. \] Substitute \( f = 100 \, \mathrm{Hz} \) and \( L = \frac{1}{5\pi} \, \mathrm{H} \): \[ X_L = 2\pi \times 100 \times \frac{1}{5\pi} = 40 \, \Omega. \] The impedance \( Z \) of the circuit is given by: \[ Z = \sqrt{R^2 + X_L^2}. \] Substitute \( R = 30 \, \Omega \) and \( X_L = 40 \, \Omega \): \[ Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500}. \] Simplify: \[ Z = 50 \, \Omega. \] The current \( I \) is calculated using Ohm's Law: \[ I = \frac{V}{Z}. \] Substitute \( V = 200 \, \mathrm{V} \) and \( Z = 50 \, \Omega \): \[ I = \frac{200}{50} = 4 \, \mathrm{A}. \] Thus, the current flowing through the coil is: \[ I = 4 \, \mathrm{A}. \]