Three Simple Circuits — Question 2

Question 2

A capacitor current leads because it tracks the voltage slope

An ideal capacitor of $C=4.00\,\mu F$ has $v=20.0\cos(500t)$ volts, using passive current direction.

1. Find i(t), rms current, and capacitive reactance.

2. Find capacitor energy when voltage is zero and when current is zero.

Question 2 - Solution

A capacitor current leads because it tracks the voltage slope

(a) Method and calculation

$$i=C\dot v=-\omega CV_p\sin\omega t.$$$$\boxed{i=-.0400\sin(500t)\,A,\quad I_{\mathrm{rms}}=.0282843\,A,\quad X_C=1/(\omega C)=500\,\Omega}.$$

(b) Interpretation and further calculation

$$U_C=\tfrac12Cv^2,\quad v=0\Rightarrow U_C=0.$$$$\boxed{i=0:\quad U_C=\tfrac12CV_p^2=.800\,mJ}.$$

Physical check

The current reaches its extrema at voltage zeros; a large current need not mean large stored electric energy.