Given:

Inductance, L = 0.5H

Resistance, R = \( 100\Omega \)

RMS voltage, \( V_{rms} = 240V \)

Frequency, f = 50Hz

Now, angular frequency, \( \omega = 2\Pi f \)

\( \omega = 2\Pi * 50 = 100\Pi rad/s \)

Inductive reactance, \( X_{L} = \omega _{L} \)

\( X_{L} = 100\Pi * 0.5 = 50\Pi \)

The formula for impedance for LR circuit;

\( z = \sqrt{X_{L}^{2} + R^{2}} \)

Therefore, z = \( \sqrt{(50\Pi )^{2} + (100)^{2}} = 186.209\Omega \)

(A) Find The Maximum Current In The Circuit

\( v_{rms} = z.I_{rms} = Ohm’s law for AC \)

\( v_{rms} = z.I_{rms} \)

= \( 240 = 186.299 * I_{rms} \)

= \( I_{rms} = (\frac{240}{186.209})A = 1.2888A \)

As \( I_{rms} = (\frac{I_{max}}{\sqrt{2}}) \)

= \( I_{max} = \sqrt{2}I_{rms} = (\sqrt{2} * 1.2888)A = 1.8226A \)

Therefore, the maximum current in the circuit, \( I_{max} = 1.8226A \)

(B) What Is The Time Lag Between Voltage Maximum And Current Maximum?

\( cos \phi = \frac{R}{z} = (\frac{100}{186.209}) \)

= \( \phi = cos^{-1} (\frac{100}{186.209}) = 1 radian \)

Therefore, t time lag, t = \( \frac{\phi}{w} = (\frac{1}{100\Pi})sec \)

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