Question

An inductor of self inductance 1 H connected in series with a resistor of 100 $ \Omega $ and an AC supply of 10 V, 50 Hz. Maximum current flowing in the circuit is:

Hint:

The current in an RL circuit is determined by the impedance, which depends on the resistance and the inductive reactance.

Answer 1

Correct Answer: 1

Impedance of the circuit:

\[ Z = \sqrt{R^2 + (X_L)^2} = \sqrt{R^2 + (\omega L)^2} \]

\[ Z = \sqrt{(100\pi)^2 + (2\pi \times 50 \times 1)^2} \]

\[ Z = \sqrt{(100\pi)^2 + (100\pi)^2} \]

\[ Z = \sqrt{2} \times 100\pi \]

RMS Current:

\[ I_{rms} = \frac{V}{Z} = \frac{100\pi}{\sqrt{2} \times 100\pi} = \frac{1}{\sqrt{2}} \]

Maximum Current:

\[ I_{max} = \sqrt{2} I_{rms} = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1 \, \text{Ampere} \]

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Final Answers:

• Impedance of the circuit: \( Z = \sqrt{2} \times 100\pi \, \Omega \)
• RMS Current: \( I_{rms} = \frac{1}{\sqrt{2}} \, \text{A} \)
• Maximum Current: \( I_{max} = 1 \, \text{A} \)

Answer 2

The impedance of an R-L circuit is determined using the formula:

\[ Z = \sqrt{R^2 + (X_L)^2} = \sqrt{R^2 + (\omega L)^2} \]

Substitute the given values:

\[ R = 100\pi, \quad \omega = 2\pi f = 2\pi \times 50, \quad L = 1 \, \text{H} \]

Therefore,

\[ Z = \sqrt{(100\pi)^2 + (2\pi \times 50 \times 1)^2} \]

On simplification,

\[ Z = \sqrt{(100\pi)^2 + (100\pi)^2} = 100\pi\sqrt{2} \]

Calculation of RMS Current:

The RMS current is given by:

\[ I_{rms} = \frac{V}{Z} = \frac{100\pi}{100\pi\sqrt{2}} = \frac{1}{\sqrt{2}} \]

Calculation of Maximum (Peak) Current:

Since \( I_{max} = \sqrt{2} \, I_{rms} \), we have:

\[ I_{max} = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1 \, \text{A} \]

Hence,

• The maximum current is \( I_{max} = 1 \, \text{A} \)