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

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

Consider that an alternating emf $\displaystyle E={{E}_{0}}sin\omega t$ is applied to a series RC circuit. Since both components are in series, the current flowing through them at any instant will be the same. Let the current in the circuit at a certain instant be $I$ ; then at that instant, the potential differences across $R$ and $C$ are V_R and V_C, respectively, where :

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

Similarly, the maximum potential differences across R and C are :

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

As shown in figure (b), V_0R will be in phase with the current, and V_0C will lag the current by 90°. Therefore, V_0R and V_0C will be perpendicular to each other. The resultant voltage E₀ is :

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

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

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

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

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

By comparing the above equation with Ohm’s law, we find that $\displaystyle \sqrt{{{\left( R \right)}^{2}}+{{\left( {{X}_{C}} \right)}^{2}}}$ represents the opposition offered to the flow of current by a series RC circuit. This is called the impedance (Z) of the circuit. Therefore,

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

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

$\displaystyle tan\phi =\frac{{{V}_{0C}}}{{{V}_{0R}}}=\frac{-{{I}_{0}}{{X}_{C}}}{{{I}_{0}}R}=\frac{-{{X}_{C}}}{R}=-\frac{{}^{1}\!\!\diagup\!\!{}_{\omega C}\;}{R}=\frac{-1}{\omega RC}$ …..(3)

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

Here, the negative sign indicates that the resultant voltage E₀ lags behind the current I₀ in phase.

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

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

Next Topic :- Comparative Study Of Various Alternating Current Circuits

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

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