Question

(1)A coil of inductance $0.50H$and resistance $100\Omega$ is connected to a $240V,50Hz$ , ac supply.
What is the maximum current in the coil?

(2)A coil of inductance $0.50H$and resistance $100\Omega$ is connected to a $240V,50Hz$ , ac supply.
What is the time lag between the voltage maximum and the current maximum?

Answer 1

(1)

Step 1: Find circuit impedance

Given,$V_{rms}=240V,f=50Hz$
$L=0.50H,R=100\Omega$

The circuit impedance is given by $Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+(\omega L{)}^2}$
$Z=\sqrt{R^2+(2\pi fL{)}^2}$

$Z=\sqrt{(100{)}^2+(2\pi \times 50\times 0.5{)}^2}$

$Z=\sqrt{(100{)}^2+(157.07{)}^2}$

$Z=189.14\Omega$

Step 2: Find maximum current

For an LR circuit if, $V=V_0\sin \omega t$

In LR circuit, current $\left(I\right)$ lags from voltage $\left(V\right)$.

$I=I_0\sin (\omega t-\varphi )$

$I=\frac{V_0}{Z}\sin (\omega t-\varphi )$

Then the maximum current is,

$I=\frac{V_0}{Z}$

Where, $V_0=\sqrt{2}{V}_{rms}=240\sqrt{2}V$

Hence from equation (i),

$I_0=\frac{240\sqrt{2}}{186.14}=1.82A$

Final Answer: $1.82A$

(2)Step 1: Calculate phase angle

Given,,$V_{rms}=240V,f=50Hz$

$L=0.50H,R=100\Omega$

Inductive reactance,
$X_L=\omega L=2\pi fL$

$=2\pi \times 50\times 0.5=157.07\Omega$
Now, from phasor diagram

$\tan \varphi =\frac{X_L}{R}$

$\varphi ={\tan }^{-1}\left(\frac{157.07}{100}\right)$

$\varphi ={57.5}^{\circ }$

Step 2: Find the time lag between the voltage maximum and the current maximum

For an LR circuit let, $V=V_0\cos \omega t$

Voltage is maximum when $t$ is zero.

Current in a LR circuit, $I=I_0\cos (\omega t-\varphi )$

Current is maximum when

$\omega t-\varphi =0$
$\Rightarrow =t=\frac{\varphi }{\omega }$ ....(i)

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

$t=3.2\times {10}^{-3}s$

Final Answer: $3.2\times {10}^{-3}s$