A coil has resistance $30$ ohm and inductive reactance $20$ ohm at $50 \,Hz$ frequency. If an ac source, of $200$ volt, $100\, Hz$, is connected across the coil, the current in the coil will be

Question

In an a.c. circuit, voltage and current are given by: \[ V = 100 \sin(100t) \, \text{V} \quad \text{and} \quad I = 100 \sin\left(100t + \frac{\pi}{3}\right) \, \text{mA}. \] The average power dissipated in one cycle is:

Answer 1

To determine the current in the coil when an AC source of 200 V and 100 Hz is connected, we need to calculate the impedance of the coil at 100 Hz. The coil has the following given parameters:

Resistance, R = 30 \, \text{ohm}
Inductive reactance at 50 Hz, X_{L_1} = 20 \, \text{ohm}
Frequency at which reactance is given, f_1 = 50 \, \text{Hz}

First, we calculate the inductance L of the coil using the formula for inductive reactance:

X_{L_1} = 2\pi f_1 L

Rearranging to solve for L:

L = \frac{X_{L_1}}{2\pi f_1} = \frac{20}{2\pi \times 50} \, \text{henry} = \frac{1}{5\pi} \, \text{henry}

Next, calculate the inductive reactance at 100 Hz using:

X_{L_2} = 2\pi f_2 L

Where f_2 = 100 \, \text{Hz}. Substitute the values:

X_{L_2} = 2\pi \times 100 \times \frac{1}{5\pi} = 40 \, \text{ohm}

Now, calculate the impedance Z of the coil at 100 Hz:

Z = \sqrt{R^2 + X_{L_2}^2}

Substituting the values, we get:

Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \text{ohm}

Finally, calculate the current I using Ohm's Law for AC circuits:

I = \frac{V}{Z} = \frac{200}{50} = 4 \, \text{A}

Therefore, the current in the coil is 4.0 A which matches the correct answer.

Answer 2