Question

100 volt AC source of angular frequency 500 rad/s is connected to a LCR circuit with L 0.8 H, C 5 uF and R 10 , all connected in series The potential difference across the resistance is

Answer 1

Answer:

The potential difference across the resistance = 100 V.

Explanation:

Given:

• AC supply voltage, V = 100 V.
• Angular frequency of the supply, $\omega$ = 500 rad/s.
• Inductance of the inductor, L = 0.8 H.
• Capacitance of the capacitor, C = 5 $\rm \mu F=5\times 10^{-6}\ F$.
• Resistance of the resistor, R = 10 $\Omega$.

The effective impedence provided by the inductor, called inductive reactance, is given by

$\rm X_L = \omega L=500\times 0.8=400\ \Omega.$

The effective impedence provided by the capacitor, called capacitive reactance, is given by

$\rm X_C = \dfrac{1}{\omega C}=\dfrac 1{500\times 5\times 10^{-6}}=400\ \Omega.$

The effective impendence of series LCR circuit is given by

$\rm Z=\sqrt{R^2+(X_L-X_C)^2}\\=\sqrt{10^2+(400-400)^2}\\=10\ \Omega.$

Using Ohm's law, the current flowing through the supply is given by

$\rm I = \dfrac{V}{Z}=\dfrac{100}{10}=10\ A.$

Thus, the potential difference across the resistance is given by

$\rm V_R = I\times R=10\times 10 = 100\ V.$