Three Simple Circuits — Question 4

Question 4

A quarter-cycle timing measurement distinguishes capacitive from inductive response

At f=200 Hz, an ideal single reactive element draws rms current .120 A from rms voltage 30.0 V. Its current maximum occurs 1.25 ms before the next voltage maximum.

1. Identify the element from the phase lead and find its value.

2. Predict rms current when frequency doubles with voltage fixed.

Question 4 - Solution

A quarter-cycle timing measurement distinguishes capacitive from inductive response

(a) Method and calculation

$$T=1/f=5.00\,ms,\quad\Delta t=T/4\Rightarrow i\ \text{leads}\ v\ \text{by}\ \pi/2.$$$$\boxed{\text{capacitor},\quad C=\frac{I_{\mathrm{rms}}}{2\pi fV_{\mathrm{rms}}}=3.18310\,\mu F}.$$

(b) Interpretation and further calculation

$$I_{\mathrm{rms}}=\omega CV_{\mathrm{rms}}\propto f.$$$$\boxed{I_{\mathrm{rms}}(400\,Hz)=.240\,A}.$$

Physical check

The timing direction matters: the same magnitude ratio with a lag would instead identify an inductor.