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

# 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

1. A

2.0 A

2. B

4.0 A

3. C

8.0 A

4. D

$\frac{20}{\sqrt{13}}\text{A}$

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### Solution:

Understanding AC Circuit Behavior:

In an alternating current (AC) circuit, the behavior of components like resistors and inductors can change with frequency. In this case, we have a coil with a resistance (R) of 30 ohms and an inductive reactance (XL) of 20 ohms at a frequency (f) of 50 Hz.

Inductive Reactance (XL):

XL represents the resistance that an inductor offers to changes in current flow. It depends on both the frequency (f) and the coil's inductance (L). Mathematically, XL is given by XL = ωL, where ω is the angular frequency (2πf).

At 50 Hz, XL is calculated as follows:

XL = 2πfL = 2π × 50 × L

This gives us XL = 100πL ohms. (Equation 1)

Changing Frequency:

Now, let's consider what happens when we change the frequency of the AC source to 100 Hz. At this new frequency, the inductive reactance XL' is calculated similarly:

XL' = 2πf'L = 2π × 100 × L

XL' = 200πL ohms.

Comparing this with Equation 1, we notice that XL' is now twice the value of XL. This means that when we increase the frequency, the inductive reactance also increases.

Impedance (Z):

Impedance (Z) is a measure of the total opposition to the flow of current in an AC circuit. It depends on both resistance (R) and reactance (XL').

Z = √(R² + XL'²)

Substituting the given values:

Z = √(30² + (200πL)²)

Calculating this expression, we find Z to be 50 ohms.

Current (I):

Finally, we can determine the current (I) using Ohm's law, which relates voltage (V) and impedance (Z):

I = V / Z

Given that the voltage (V) is 200 volts and the impedance (Z) is 50 ohms, we find:

I = 200 / 50

I = 4 amperes (A).

So, when the frequency of the AC source is changed to 100 Hz, the current in the coil will be 4 amperes.

The Correct Option is B 4.0 A

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