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NEET PHYSICSMedium

A coil has resistance 30 \Omega and inductive reactance 20 \Omega 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:

A

4.0 A

B

8.0 A

C

20√13 A

D

2.0 A

Step-by-Step Solution

The inductive reactance (XLX_L) is directly proportional to the frequency of the AC source, given by XL=ωL=2πνLX_L = \omega L = 2\pi \nu L .

  1. Calculate new reactance: At 50 Hz, XL=20 ΩX_L = 20 \, \Omega. Since the new frequency is 100 Hz (double the initial frequency), the new reactance XL′X_L' becomes 2×20 Ω=40 Ω2 \times 20 \, \Omega = 40 \, \Omega.
  2. Calculate Impedance: The coil consists of a resistor and an inductor in series. The impedance ZZ is calculated as Z=R2+(XL′)2=302+402=900+1600=2500=50 ΩZ = \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 V/50 Ω=4.0 AI = V / Z = 200 \text{ V} / 50 \, \Omega = 4.0 \text{ A}.
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