Question

$FIGUREEX32.9$shows voltage and current graphs for a capacitor.

a. What is the emf frequency $f$?

b. What is the value of the capacitance $C$?

Short answer

(a) The emf frequency is $f=50Hz$.

(b) The value of the capacitance is$C=0.47\times {10}^{-5}F$.

Step-by-step solution

Part(a) Step 1: Given information

We have been given that the voltage and current graphs for a capacitor in$FIGUREEX32.9$.

We need to find the emf frequency $f$.

Part(a) Step 2: Simplify

As the graph for voltage repeats after $0.02s$.

Therefore, the Time period is $T=0.02s$.

We know the formula, localid="1649884330912" $f=\frac{1}{T}$ (Let this equation be ($1$))

Substitute the value of T in equation ($1$),

$f=\frac{1}{0.02}$

On simplifying,

$f=50Hz$

Part(b) Step 1: Given information

We have been given that the voltage and current graphs for a capacitor in$FIGUREEX32.9$.

We need to find the the value of the capacitance localid="1650106009664" $C$.

Part(b) Step 2: Simplify

Given values: $V_c=10V$

$I_c=15\times {10}^{-3}A$

$f=50Hz$ (from Part(a))

$\omega =\left(2\pi \right)f=\left(2\pi \right)\times 50$

We know the formula, $I_c=V_c\omega C$ (Let this equation be ($2$))

Divide both sides of equation ($2$) by $V_c\omega$,

localid="1649928007971" $\frac{I_c}{V_c\omega }=\frac{V_c\omega C}{V_c\omega }$

$C=\frac{I_c}{V_c\omega }$ (Let this equation be ($3$))

Substitute the values of $V_c,I_c,\omega$in equation ($3$),

$C=\frac{15\times {10}^{-3}}{10\times \left(2\pi \right)\times 50}$

On simplifying,

$C=0.47\times {10}^{-5}F$