A coil has resistance 30 Ω and inductive reactance 20 Ω at 50 Hz frequency. If an AC source of 200 V, 100 Hz,

Question

A coil has resistance 30 Ω and inductive reactance 20 Ω at 50 Hz frequency. If an AC source of 200 V, 100 Hz, is connected across the coil, the current in the coil will be:

Accepted Answer

The inductive reactance ($X_L$) is directly proportional to the frequency of the AC source, given by $X_L = \omega L = 2\pi 
u L$ . 
1. **Calculate new reactance:** At 50 Hz, $X_L = 20 \, \Omega$. Since the new frequency is 100 Hz (double the initial frequency), the new reactance $X_L'$ becomes $2 	imes 20 \, \Omega = 40 \, \Omega$. 
2. **Calculate Impedance:** The coil consists of a resistor and an inductor in series. The impedance $Z$ is calculated as $Z = \sqrt{R^2 + (X_L')^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \Omega$ . 
3. **Calculate Current:** Using Ohm's law for AC circuits, $I = V / Z = 200 	ext{ V} / 50 \, \Omega = 4.0 	ext{ A}$.

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