Question

a. What is the peak current supplied by the emf in FIGURE$P32.45$?

b. What is the peak voltage across the $3.0mF$capacitor?

Short answer

a. The peak current supplied by the emf in FIGURE $P32.45$is $25mA$

b. The peak voltage across the $3.0mF$capacitor is$6.7V$

Step-by-step solution

Part(a) Step 1: Given information 

We are given that the frequency $200Hz$, localid="1649935325324" ${\epsilon }_0=10V$, $C=3\mu F$and $C_{eq}=2\mu F$

We need to find peak current supplied by the emf in FIGURE$P32.45$

Part(a) Step 2: Simplify 

The peak current can be calculated by $I=\frac{V}{X_C},$We can calculate and by

$V={\epsilon }_0,X_C=\frac{1}{2\mathrm{\pi fC}}$, now by putting these value we can get ,

$I=\frac{{\epsilon }_0}{{\displaystyle \frac{1}{2\mathrm{\pi fC}}}}=2{\mathrm{\pi fC\epsilon }}_0$

Finally putting all the values in above equation , we get

$I=2{\mathrm{\pi fC\epsilon }}_0=25\mathrm{mA}$

Part(b) step 1: Given information 

We are given that frequency $200Hz$,${\epsilon }_0=10V$,$C=3\mu F$and $C_{eq}=2\mu F$

We have to peak voltage in $3.0mF$capacitor.

Part(b) Step 2: Simplify 

We can find peak voltage through the given formula ,$V_C=I\times X_C$

Where , $I=2{\mathrm{\pi fC}}_{\mathrm{eq}}{\mathrm{\epsilon }}_0,{\mathrm{X}}_{\mathrm{C}}=\frac{1}{2\mathrm{\pi fC}}$

Putting the values , we get

$V_C=\frac{C_{eq}}{C}{\epsilon }_0$

Substituting the values ,$V_C=\frac{C_{eq}}{C}{\epsilon }_0=\frac{2}{3}\times 10=6.7V$

$$\begin{gathered} V_C=I·X_C \\ {X}_C=\frac{1}{2\mathrm{\pi fC}},I=2{\mathrm{\pi fC}}_{\mathrm{eq}}{\mathrm{\epsilon }}_0 \end{gathered}$$