Circuit containing Capacitor and Resistor in Series (C-R Series Circuit )

Mathematical Analysis of C-R Series Circuit :
Alternating Current Circuit Containing Capacitor and Resistor in series
Let us consider, a circuit containing capacitor $C$ resistor $R$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the C-R circuit

$V=\sqrt{ V_{C} ^{2} + V^{2}_{R}} \qquad(1)$

We know that:

$V_{R} = iR$
$V_{C} = iX_{C}$

So from equation $(1)$

$V=\sqrt{\left( iX_{C} \right)^{2} + \left(iR\right)^{2}} $

$V=i\sqrt{\left( X_{C} \right)^{2} + R^{2}} $

$\frac{V}{i}=\sqrt{\left( X_{C} \right)^{2} + R^{2}} $

$Z=\sqrt{\left( X_{C} \right)^{2} + R^{2}} \qquad(2)$

Where
$Z \rightarrow$ Impedance of C-R circuit.
$X_{C} \rightarrow$ Capacitive Reactance which has value $\frac{1}{\omega C}$

So from equation $(2)$, we get

$Z=\sqrt{\left( \frac{1}{\omega C} \right)^{2} + R^{2}} \qquad(3)$

The phase of resultant voltage:
Phasor Diagram For C-R Circuit
If the phase of resultant voltage from from current is $\phi$ then

$tan \phi = \frac{X_{C} }{R} \qquad(4)$

$tan \phi = \frac{\frac{1}{\omega C}}{R} $

$tan \phi = \frac{1}{\omega C R} $

$\phi = tan^{1} \left(\frac{1}{\omega C R}\right) $

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