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

2 Answers
José Roberto Pereira · A08
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.

A08
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=1/(jomega C)#, where #j-=sqrt -1# also

Capacitive Reactance #X_C -=(omegaC)^-1#

Now in a parallel #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.

#1/Z_(eq)=1/Z_R+1/Z_C=1/R+jomega C=1/R+j/(X_C)#
#implies Z_(eq)=(RX_C)/(X_C+jR)#

The phase angle between the applied voltage and total current is given by
#theta=tan^-1 (R/X_C)#
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.

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