41. Power Factor Correction for Lamp
Step 1: Analyze Lamp Branch
The lamp assembly consists of a resistance $R$ (tube) and an inductance $L$ (choke) in series. Same current $I$ flows through both.
Given: Voltage on choke ($V_L$) is $\eta$ times voltage on tube ($V_R$).
$$ V_L = \eta V_R \implies I X_L = \eta I R \implies \omega L = \eta R $$ $$ R = \frac{\omega L}{\eta} $$Step 2: Condition for Unity Power Factor
To make the Power Factor unity, the net reactive component of the circuit must be zero. We connect a capacitor $C$ in parallel. The inductive susceptance of the lamp branch must be canceled by the capacitive susceptance.
Admittance of Lamp Branch ($Y_L$):
$$ Z_{lamp} = R + j\omega L $$ $$ Y_{lamp} = \frac{1}{R + j\omega L} = \frac{R – j\omega L}{R^2 + (\omega L)^2} $$The imaginary part (inductive susceptance) is: $B_L = \frac{-\omega L}{R^2 + \omega^2 L^2}$.
Admittance of Capacitor ($Y_C$):
$$ Y_C = j \omega C $$The imaginary part is: $B_C = \omega C$.
Step 3: Equating Susceptances
For Unity PF, $B_C + B_L = 0 \implies \omega C = \frac{\omega L}{R^2 + \omega^2 L^2}$.
$$ C = \frac{L}{R^2 + \omega^2 L^2} $$Step 4: Substitute R
Using $R = \frac{\omega L}{\eta}$:
$$ R^2 + \omega^2 L^2 = \left(\frac{\omega L}{\eta}\right)^2 + (\omega L)^2 = \omega^2 L^2 \left( \frac{1}{\eta^2} + 1 \right) = \omega^2 L^2 \left( \frac{1 + \eta^2}{\eta^2} \right) $$Substitute back into the expression for C:
$$ C = \frac{L}{\omega^2 L^2 \left( \frac{1 + \eta^2}{\eta^2} \right)} = \frac{1}{\omega^2 L} \left( \frac{\eta^2}{\eta^2 + 1} \right) $$