Question

A high-pass RC filter is connected to an AC source with a peak voltage of $10.0V$. The peak capacitor voltage is $6.0V$. What is the resistor voltage?

Short answer

The resistor voltage is $8V$.

Step-by-step solution

Given information

We have been given that,

Peak Voltage of an A.C. Circuit is, $V_0=10.0V$

And peak Capacitor Voltage is,$V_C=6.0V$

Simplify

Let Resistor Voltage$=V_R$

In RC Filter, we know that,

Peak voltage is,

$V_0=\sqrt{{V_R}^2+{V_C}^2}$

${V_0}^2={V_R}^2+{V_C}^2$

$$\begin{gathered} {10}^2={V_R}^2+6^2 \\ 100={V_R}^2+36 \\ {V_R}^2=100-36 \\ {V_R}^2=64 \\ {V}_R=\sqrt{64} \\ {V}_R=8V \end{gathered}$$

Hence, resistor voltage is $8V$.