Question

Deduce expression for the resultant potential difference, impedance and current in a L-R alternating circuit. Draw graph to show the phase difference between the current and voltage.

Answer 1

A pure resistance $R$ and a pure inductive coil of inductance $L$ as shown connected in series (fig A).

Let $V$= r.m.s. value of the applied voltage, $I=r.m.s$ value of the resultant current.

$V_R=IR-$ voltage drop across $R$ (in phase with $I$), $V_L=I.X_L-$ voltage drop across coil (ahead of $I$ by ${90}^0$).

The voltage drops are shown in voltage triangle OAB in figure B. The applied voltage $V$ is the vector sum of the two $i.e.$ $OB$.

$\therefore V=\sqrt{(V_R^2-V_L^2}=I\sqrt{R^2-X_L^2}$

The quantity $\sqrt{(R^2+X_L^2)}$ is known as the impedance $Z$ of the circuit. As seen from the impedance triangle $ABC$ (figure C), $Z^2=R^2+X_L^2$ i.e. ${Impedance}^2={Resistance}^2+{Reactance}^2$

From figure B it is clear that the applied voltage $V$ leads the current $I$ by an angle $\varphi$ such that

$tan\varphi =\frac{V_L}{V_R}=\frac{I.X_L}{I.R}=\frac{X_L}{R}=\frac{\omega L}{R}$

$\therefore \varphi =ta{n}^{-1}\frac{X_L}{R}$

The same fact is illustrated graphically in the figure D.

In other words, current $I$ lags behind the applied voltage $V$ by an angle $\varphi$

Hence, if applied voltage is given by $v=V_{m}sin\omega t$, then current equation is

$i=I_{m}sin(\omega t-\varphi )$ where $I_m=\frac{V_m}{Z}$