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A coil of inductance $0.50\;H$ and resistance $100 \;\Omega$ is connected to a $240\; V , 50\; Hz$ $ac$ supply.
$(a)$ What is the maximum current in the coil?
$(b)$ What is the time lag between the voltage maximum and the current maximum?
Solution
Inductance of the inductor, $L=0.50\, H$
Resistance of the resistor, $R =100 \,\Omega$
Potential of the supply voltage, $V=240\, V$
Frequency of the supply, $v=50 \,Hz$
$(a)$ Peak voltage is given as:
$V_{0}=\sqrt{2 V}$
$=\sqrt{2} \times 240=339.41 \,V$
Angular frequency of the supply, $\omega=2 \pi v$ $=2 \pi \times 50=100 \pi\, rad / s$
Maximum current in the circuit is given as:
$I_{0}=\frac{V_{0}}{\sqrt{R^{2}+\omega^{2} L^{2}}}$
$=\frac{339.41}{\sqrt{(100)^{2}+(100 \pi)^{2}(0.50)^{2}}}=1.82\, A$
$(b)$ Equation for voltage is given as: $V = V _{0} \cos \omega t$
Equation for current is given as: $I=I_{0} \cos (\omega t-\phi)$ Where.
$\phi=$ Phase difference between voltage and current At time, $t=0 . V = V _{0}($ voltage is maximum)
For $\omega t-\phi=0$ i.e., at $t=\frac{\phi}{\omega}$
$I=I_{0}$ (current is maximum) Hence, the time lag between maximum voltage and maximum current is $\frac{\phi}{\omega}$
$\tan \phi=\frac{\omega L}{R}$
$=\frac{2 \pi \times 50 \times 0.5}{100}=1.57$
$\phi=57.5^{\circ}=\frac{57.5 \pi}{180} \,rad$
$\omega t=\frac{57.5}{180 \times 2 \pi \times 50}$
$=3.19 \times 10^{-3}\, s$
$=3.2 \,ms$
Now, phase angle $\phi$ is given by the relation, Hence, the time lag between maximum voltage and maximum current is $3.2 \,ms$.
