Solution Electrical Numerical 13

Numerical 13: A coil having a resistance of 10 Ohm, and an inductance of 0.15 H is connected in series with a capacitor across a 100V, 50Hz supply. If the current and the voltage are in phase what will be the value of the current in the circuit and the voltage drop across the coil? 

Solution: Given, Coil resistance (R) = 10ᘯ

Inductance = 0.15 H

Voltage = 100 V

Frequency(f) = 50 Hz

We know $\displaystyle \mathrm{I = \frac{V}{R}}$ $\displaystyle \mathrm{= \frac{100}{10}=10\,A}$ 

This current 10 A will flow through Coil and Capacitor.

Impedance of coil = $\displaystyle \mathrm{Z=\sqrt{(R)^{2}+(X_L)^2}}$ 

$\displaystyle \mathrm{Z=\sqrt{(R)^{2}+(2\pi fL)^2}}$ 

$\displaystyle \mathrm{Z=\sqrt{(10)^{2}+(2\times 3.14\times 50\times 0.15)^2}}$ =48.1498 ᘯ

Now Voltage across coil is given,

$\displaystyle \mathrm{V_1=IZ\, =10\times 48.1498\ = 481.498\ V}$