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 (Identify \(\omega\)): The applied EMF \(V = 200\sin 100t\) tells us the drive frequency directly: \(\omega = 100\ \text{rad/s}\) (the coefficient of \(t\)).

Step 2 (Two opposing reactances): The coil and capacitor push the phase in opposite directions. Compute each: \(X_L = \omega L = 100(10) = 1000\ \Omega\) and \(X_C = 1/(\omega C) = 1/(100\cdot 20\times10^{-6}) = 500\ \Omega\).

Step 3 (Reactance triangle): Represent \(R\) along the horizontal and the net reactance \((X_L - X_C) = 500\ \Omega\) along the vertical. Because \(R = 500\ \Omega\) and the net reactance is also \(500\ \Omega\), this is an isosceles right triangle.

Step 4 (Impedance = hypotenuse): For an isosceles right triangle the hypotenuse is \(\sqrt2\) times a leg:
\[ Z = R\sqrt2 = 500\sqrt2 \approx 707\ \Omega \]

Step 5 (Angle of the triangle): Equal legs mean a \(45^\circ\) angle, so the phase difference is \(\phi = 45^\circ\), with voltage ahead of current (inductive side larger).

Step 6 (Power factor = adjacent/hypotenuse):
\[ \cos\phi = \cos 45^\circ = \frac{1}{\sqrt2} \approx 0.707 \]
All three answers follow from one right triangle.
\[\boxed{Z \approx 707\ \Omega,\ \ \cos\phi = 0.707,\ \ \phi = 45^\circ}\]