17.11 AC Through RL-Series Circuit

17.11 AC Through RL-Series Circuit

Consider an AC circuit containing a resistor $R$ and inductor $L$ connected in series with an AC source $V$ as shown in Fig. 17.10(a). As $R$ and $L$ are in series, therefore, same current $I$ flows through both $R$ and $L$. $I = I_{o} \sin\theta$

(Fig. 17.10 (a): RL series circuit connected with A.C. supply)

(Fig. 17.10 (b): Voltage and Current waveform in of RL series circuit.)

Let rms value of current through $R$ is $I$. Then rms value of voltage $V_{R}$ across $R$ will be: $V_{R} = I_{\mathrm{rms}} R$ $V$ will be in phase with the current $I$ in the circuit, as $R$ is shown vectorially by line OA in Fig. 17.10(c). The value of rms voltage $V_{L}$ across $L$ will be: $V_{L} = I_{\mathrm{rms}} X_{L}$ Voltage $V$ across $L$ is ahead of current $I$ through $L$ by $90^{\circ}$ as shown in Fig. 17.10 (b) and vectorially by line OB in Fig. 17.10(c). The resultant voltage $V$ will be the vector sum of $V_{R}$ and $V_{L}$ shown vectorially by line OP, thus in Fig. 17.10(c): $V_{\mathrm{rms}} = \sqrt{V_{R}^{2} + V_{L}^{2}}$ $V_{\mathrm{rms}} = \sqrt{(I_{\mathrm{rms}} R^{2}) + (I_{\mathrm{rms}} X_{L}^{2})}$ As $X_{L} = \omega L$ , so $V_{\mathrm{rms}} = I_{\mathrm{rms}}\sqrt{R^{2} + (\omega L)^{2}}$ Impedance

Angle $\theta = \tan^{-1}\left(\dfrac{\omega L}{R}\right)$ The angle $\theta$ which $Z$ makes with $R$ line, gives the phase difference between the applied voltage and current.

(Fig. 17.10 (c): Vector diagram of $V_{R}$ and $V_{L}$ in of RL series circuit.)

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