Question

An inductance and a capacitance are connected in series with a source of alternating e.m.f. Derive an expression for resultant voltage, impedance and phase difference between current and voltage in alternating circuit.

Answer 1

Ohmic resistance R and inductance L are connected in series with an alternating e.m.f. source S. The e.m.f. V applied by the source is.
$V=V_{o}sin\omega t$
If at any instant, the current in the circuit be I due to which potential difference across the inductance L is
$V_L=I{X}_L$
where $X_L$ is the inductive reactance
Similarly potential difference across resistance R is,
$V_R=IR$
But I and $V_R$ are in same phase and $V_L$ is ahead by $V_R$ by ${90}^o$, thus
(1) Resultant voltage $V=\sqrt{V_L^2+V_R^2}$
or $V=\sqrt{(I{X}_L{)}^2+(IR{)}^2}=\sqrt{I^2{X}_L^2+I^2{R}^2}=I\sqrt{X_L^2+R^2}$
or $V=I\sqrt{{\omega }^2{L}^2+R^2}$ .......(i)
$[\because X_L=\omega L]$
(2) Impedence $Z=\frac{V}{I}=\sqrt{{\omega }^2{L}^2+R^2}$
Thus $Z=\sqrt{{\omega }^2{L}^2+R^2}$ $[\because ImpedenceZ=\frac{V}{I}]$
(3) Current $I=\frac{V_o}{Z}$ $[\because CurrentI=\frac{voltage{V}_o}{ImpedanceZ}]$
or $I=\frac{V_o}{\sqrt{{\omega }^2{L}^2+R^2}}$
(4) Phase difference-(between resultant potential difference V and current I)
$tan\Phi =\frac{V_L}{V_R}=\frac{I{X}_L}{IR}=\frac{X_L}{R}=\frac{\omega L}{R}$
Thus $\Phi =ta{n}^{-1}\left(\frac{\omega L}{E}\right)$.