Question

A resistance and inductance 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

As per figure resistance R and L inductance are connected in series to an a.c. source.
At any instant the a.c. voltage is given by the equation.
$V=V_{0}sin\omega t$
If I current flows through circuit
PD across R will be $=V_R=IR$
PD across L will be $=V_L=I{X}_L$
Now, $V_R$ and I are in the same phase but $V_L$ is leading I by a phase difference of 90${}^o$ thus angle between $V_L$ and $V_R$ is ${90}^o$.
Resultant of $V_R$ and $V_L$ is V
$V^2=V_R^2+V_L^2$
$V^2=(IR{)}^2+(I{X}_L{)}^2$
$V^2=I^2[R^2+X_L{]}^2$
$\frac{V^2}{I^2}=R^2+X_L^2\Rightarrow \frac{V}{I}=Z$ called impedance
$Z^2=R^2+X_L^2\Rightarrow Z=\sqrt{R^2+X_L^2}$
$X_L=L\omega$
$Z=\sqrt{R^2+L^2{\omega }^2}$
V is leading current I flows in circuit
$I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+L^2{\omega }^2}}$
If phase difference between V and I $\varphi$, then,
$tan\varphi =\frac{V_L}{V_R}=\frac{I{X}_L}{IR}=\frac{X_L}{R}$
Phase difference
$\varphi =ta{n}^{-1}\frac{X_L}{R}$