(a) Explain the terms reactance and impedance in an a.c circuit. (b) A source of e.m.f. 240 v and frequency 50 Hz is connected to a resistor, an inductor an...

(a) Reactance and impedance

Reactance is the opposition offered to alternating current by an inductor or a capacitor. It is measured in ohms, \(\Omega\).

For an inductor, \(X_L=2\pi fL\), while for a capacitor, \(X_C=\dfrac{1}{2\pi fC}\).

Impedance is the total opposition offered to alternating current by a circuit containing resistance and reactance. For a series RLC circuit,

\[Z=\sqrt{R^2+(X_L-X_C)^2}.\]

(b) Vector diagram

Take the current, and hence \(V_R\), as the horizontal reference. \(V_L\) leads the current by \(90^\circ\), while \(V_C\) lags the current by \(90^\circ\).

From the voltage phasor diagram,

\[V^2=V_R^2+(V_L-V_C)^2.\]

Given \(V=240\text{ V}\), \(V_R=140\text{ V}\), and \(V_L=50\text{ V}\),

\[240^2=140^2+(50-V_C)^2\]

\[(50-V_C)^2=57600-19600=38000\]

\[50-V_C=-\sqrt{38000}=-194.94\]

\[V_C=244.94\text{ V}.\]

(i) Potential difference across the capacitor:

\[\boxed{V_C\approx245\text{ V}}\]

(ii) Capacitance of the capacitor

\[X_C=\frac{V_C}{I}=\frac{244.94}{10}=24.49\ \Omega\]

\[C=\frac{1}{2\pi fX_C}=\frac{1}{2\pi\times50\times24.49}\]

\[\boxed{C=1.30\times10^{-4}\text{ F}=130\ \mu\text{F}}\]

(iii) Inductance of the inductor

\[X_L=\frac{V_L}{I}=\frac{50}{10}=5.0\ \Omega\]

\[L=\frac{X_L}{2\pi f}=\frac{5.0}{2\pi\times50}\]

\[\boxed{L=1.59\times10^{-2}\text{ H}\approx0.016\text{ H}}\]