# In parallel RC circuits, is it true that the larger the value of C, the larger the phase angle?

Oct 15, 2015

Yes.

#### Explanation:

The larger value of C, the smaller is its reactance Xc and the larger is the current Ic.

As the angle is = arctg (Ic/Ir),
the larger Ic the larger the angle.

Jan 25, 2016

Yes.

#### Explanation:

When an AC voltage having frequency $\omega$ is applied across a capacitor of capacity $C$, the impedance ${Z}_{C}$ is given by

${Z}_{C} = \frac{1}{j \omega C}$, where $j \equiv \sqrt{-} 1$ also

Capacitive Reactance ${X}_{C} \equiv {\left(\omega C\right)}^{-} 1$

Now in a parallel $R C$ circuit as shown in the figure

The total impedance is calculated by adding individual impedance in parallel.

$\frac{1}{Z} _ \left(e q\right) = \frac{1}{Z} _ R + \frac{1}{Z} _ C = \frac{1}{R} + j \omega C = \frac{1}{R} + \frac{j}{{X}_{C}}$
$\implies {Z}_{e q} = \frac{R {X}_{C}}{{X}_{C} + j R}$

The phase angle between the applied voltage and total current is given by
$\theta = {\tan}^{-} 1 \left(\frac{R}{X} _ C\right)$
We observe that larger value of capacitor will lead to larger phase angle, ${X}_{C}$ being in the denominator or $C$ being in the numerator.