A 100Ω resistor is connected in series with a choke coil and when a $400V, 50H_Z$ supply is applied to this combination, the voltages across the resistance and the choke coil are 200Vand 300V respectively. Also, calculate the power factor of the choke coil and power factor of the circuit. -

$R = 100Ω \\ V = 400V \\ f=50H_Z \\ V_R=200V \\ V_{coil}=300V \\ \text{Power consumed by choke coil:} \\ I= \dfrac{V_R}{R}= \dfrac{200}{100}=2A \\ Z_{coil}= \dfrac{V_{coil}}{I}= \dfrac{300}{2}=150Ω \\ \sqrt{r^2+X_L^2 }=150 \\ r^2+X_L^2=22500……………..(1) \\ Z= \dfrac{V}{I} =\dfrac{400}{2}=200Ω \\ \bar{Z} = ( R+V)+jX_L \\ Z= \sqrt{( R+V)^2+X_L^2}=200 \\ (100+r)^2+ X_L^2=40000………………(2) \\ \text{Subtract (1) from (2)} \\ (100+r)^2- r^2 =17500 \\ 10000+20r+r^2- r^2=17500 \\ 200r=17500-10000 \\ 200r=7500 \\ r=3.75Ω \\ \text{Substituting the value of r in equation (1)} \\ (37.5)^2 + X_L^2=22500 \\ X_L^2=21093.75 \\ X_L=145.24 Ω \\ P_{coil}= I^2 r \\ =2^2×37.5 \\ =50W \\ \text{Power factor of the choke coil} \\ pf_{coil}=\dfrac{V}{Z_{coil}} =\dfrac{37.5}{150}=0.25 (lag) \\ \text{Power factor of the circuit} \\ pf_{circuit}=\dfrac{R+r}{Z}= \dfrac{1000+37.5}{200}=0.678 (lag)$