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Series LR Circuit | A Series LR Circuit Connected With An AC Source

Series LR Circuit | A Series LR Circuit Connected With An AC Source :- In the figure (a) below, a resistor R and an inductor L are connected in series with an AC voltage source.

Consider a series LR circuit supplied with an alternating emf $\displaystyle E={{E}_{0}}sin\omega t$ . Since both components are connected in series, the current flowing through them at any instant will be the same. Let the current in the circuit at some instant be $I$ . Then, at that instant, the potential differences across $R$ and $L$ are V_R and V_L, respectively, where :

$\displaystyle {{V}_{R}}=IR\text{ }\!\!\And\!\!\text{ }{{V}_{L}}=I{{X}_{L}}$

Similarly, the maximum values of the potential differences across R and L are :

$\displaystyle {{V}_{0R}}={{I}_{0}}R\text{ }\!\!\And\!\!\text{ }{{V}_{0L}}={{I}_{0}}{{X}_{L}}$

As shown in figure (b), V_0R will be in phase with the current, and V_0L will lead the current by 90°. Therefore, V_0R and V_0L will be perpendicular to each other. The resultant voltage E₀ is :

$\displaystyle {{E}_{0}}=\sqrt{V_{0R}^{2}+V_{0L}^{2}}$

$\displaystyle \Rightarrow {{E}_{0}}=\sqrt{{{\left( {{I}_{0}}R \right)}^{2}}+{{\left( {{I}_{0}}{{X}_{L}} \right)}^{2}}}$

$\displaystyle \Rightarrow {{E}_{0}}={{I}_{0}}\sqrt{{{\left( R \right)}^{2}}+{{\left( {{X}_{L}} \right)}^{2}}}$

Thus, the maximum value of the current flowing in the circuit is I₀:

$\displaystyle \therefore {{I}_{0}}=\frac{{{E}_{0}}}{\sqrt{{{\left( R \right)}^{2}}+{{\left( {{X}_{L}} \right)}^{2}}}}$ …..(1)

By comparing the above equation with Ohm’s law, we find that $\displaystyle \sqrt{{{\left( R \right)}^{2}}+{{\left( {{X}_{L}} \right)}^{2}}}$ represents the opposition produced in the path of current by a series LR circuit. This is called the impedance (Z) of the circuit. Hence,

$\displaystyle Z=\sqrt{{{\left( R \right)}^{2}}+{{\left( {{X}_{L}} \right)}^{2}}}$ …..(2)

In figure (b), the resultant voltage E₀ leads the current by a phase angle Φ, where tanΦ is :

$\displaystyle tan\phi =\frac{{{V}_{0L}}}{{{V}_{0R}}}=\frac{{{I}_{0}}{{X}_{L}}}{{{I}_{0}}R}=\frac{{{X}_{L}}}{R}=\frac{\omega L}{R}$ …..(3)

$\displaystyle \therefore \phi =ta{{n}^{-1}}\left( \frac{\omega L}{R} \right)$

The reciprocal of impedance (Z) is called susceptance (S).

$\displaystyle S=\frac{1}{Z}=\frac{1}{\sqrt{{{\left( R \right)}^{2}}+{{\left( {{X}_{L}} \right)}^{2}}}}=\frac{1}{\sqrt{{{\left( R \right)}^{2}}+{{\omega }^{2}}{{L}^{2}}}}$

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Series LR Circuit | A Series LR Circuit Connected With An AC Source

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