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

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

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Answer 1 · 2015-10-15T22:06:10

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.

Answer 2 · 2016-01-25T23:23:24

Yes.

Explanation:

When an AC voltage having frequency $ω$ $omega$ is applied across a capacitor of capacity $C$ $C$ , the impedance $Z_{C}$ $Z_C$ is given by

$Z_{C} = \frac{1}{j ω C}$ $Z_C=1/(jomega C)$ , where $j \equiv \sqrt{-} 1$ $j-=sqrt -1$ also

Capacitive Reactance $X_{C} \equiv {(ω C)}^{- 1}$ $X_C -=(omegaC)^-1$

Now in a parallel $R C$ $RC$ circuit as shown in the figure
![http://isu.edu.tw](https://useruploads.socratic.org/Cq3Dn9gRfyfjNBo5r9At_Screen%20shot%202016-01-25%20at%202.26.18+PM.png)

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

$\frac{1}{Z_{e q}} = \frac{1}{Z_{R}} + \frac{1}{Z_{C}} = \frac{1}{R} + j ω C = \frac{1}{R} + \frac{j}{X_{C}}$ $1/Z_(eq)=1/Z_R+1/Z_C=1/R+jomega C=1/R+j/(X_C)$
$\Rightarrow Z_{e q} = \frac{R X_{C}}{X_{C} + j R}$ $implies Z_(eq)=(RX_C)/(X_C+jR)$

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

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

2 Answers

Explanation:

Explanation:

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