A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V,

Solution:

Inductance of the inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of the supply voltage, V = 240 V

Frequency of the supply, ν = 50 Hz

(a) Peak voltage is given as:

$V_{0}=\sqrt{2} V$

$=\sqrt{2} \times 240=339.41 \mathrm{~V}$

Angular frequency of the supply,

ω = 2 πν

= 2π × 50 = 100 π rad/s

Maximum current in the circuit is given as:

$I_{0}=\frac{V_{0}}{\sqrt{R^{2}+\omega^{2} L^{2}}}$

$=\frac{339.41}{\sqrt{(100)^{2}+(100 \pi)^{2}(0.50)^{2}}}=1.82 \mathrm{~A}$

(b) Equation for voltage is given as:

V = V0 cos ωt

Equation for current is given as:

I = I0 cos (ωt − Φ)

Where,

Φ = Phase difference between voltage and current

At time, t = 0.

V = V0(voltage is maximum)

For $\omega t-\Phi=0$ i.e., at time $t=\frac{\phi}{\omega}$

I = I0 (current is maximum)

Hence, the time lag between maximum voltage and maximum current is $\frac{\phi}{\omega}$.

Now, phase angle Φis given by the relation,

$\tan \phi=\frac{\omega L}{R}$

$=\frac{2 \pi \times 50 \times 0.5}{100}=1.57$

$\phi=57.5^{\circ}=\frac{57.5 \pi}{180} \mathrm{rad}$

$\omega t=\frac{57.5 \pi}{180}$

$t=\frac{57.5}{180 \times 2 \pi \times 50}$

$=3.19 \times 10^{-3} \mathrm{~s}$

$=3.2 \mathrm{~ms}$

Hence, the time lag between maximum voltage and maximum current is 3.2 ms.