Question

The total power dissipated in the circuit, shown in the figure, is 1 kW.



The voltmeter, across the load, reads 200 V. The value of $K_L$ is

• A. 17.332

Answer 1

The correct option is A 17.332

Given, total power dissipated in the circuit

= 1 kW = 1000Watts

$\therefore 2^2\times 1+{10}^2\times R=1000$

$or,R=\frac{998}{100}=9.98\Omega$

Also, voltage drop across R;

$V_R=IR=10\times 9.98$

$=99.8volt$

Voltage drop across load,

$V=200volt=\sqrt{V_R^2+V_{X_L}^2}$

$\therefore$ Voltage drop across inductor,

$V_{X_L}=\sqrt{V^2-V_R^2}=\sqrt{{\left(200\right)}^2-{\left(99.8\right)}^2}$

$=173.32volt$

$Now,V_{X_L}=I{X}_{L}or{X}_L=\frac{V_{X_L}}{I}$

$=\frac{173.32}{10}=17.332\Omega$

$\therefore X_L=17.332\Omega$