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 = $$\frac{\phi}{w} = (\frac{1}{100\Pi})sec$$

Explore more such questions and answers at BYJU’S.