Question

Find out from the given circuit shown above: (i) Impedance (ii) power factor (iii) phase difference between current and voltage. The series circuit has \(L = 10\ \text{H}\), \(R = 500\ \Omega\) and \(C = 20\ \mu F\), driven by an AC source \(V = [200\sin 100t]\ \text{Volt}\).

Hint:

Read \(\omega = 100\) from \(200\sin 100t\). Find \(X_L=\omega L\), \(X_C=1/\omega C\), then \(Z=\sqrt{R^2+(X_L-X_C)^2}\), \(\cos\phi=R/Z\), \(\tan\phi=(X_L-X_C)/R\).

Answer 1

Step 1 (Read the source): Comparing \(V = 200\sin 100t\) with \(V = V_0\sin\omega t\) gives peak voltage \(V_0 = 200\ \text{V}\) and angular frequency \(\omega = 100\ \text{rad/s}\).

Step 2 (Inductive reactance):
\[ X_L = \omega L = 100\times10 = 1000\ \Omega \]

Step 3 (Capacitive reactance): \(C = 20\ \mu F = 20\times10^{-6}\ \text{F}\).
\[ X_C = \frac{1}{\omega C} = \frac{1}{100\times20\times10^{-6}} = \frac{1}{2\times10^{-3}} = 500\ \Omega \]

Step 4 (Net reactance):
\[ X_L - X_C = 1000 - 500 = 500\ \Omega \]

Step 5 (i) Impedance:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{500^2 + 500^2} = 500\sqrt{2} \approx 707\ \Omega \]

Step 6 (ii) Power factor:
\[ \cos\phi = \frac{R}{Z} = \frac{500}{500\sqrt2} = \frac{1}{\sqrt2} \approx 0.707 \]

Step 7 (iii) Phase difference:
\[ \tan\phi = \frac{X_L - X_C}{R} = \frac{500}{500} = 1 \ \Rightarrow\ \phi = 45^\circ \]
Since \(X_L > X_C\) the circuit is inductive, so the voltage leads the current by \(45^\circ\).
\[\boxed{Z \approx 707\ \Omega,\quad \cos\phi = 0.707,\quad \phi = 45^\circ}\]