Question

Figure P$32.47$shows a parallel RC circuit.

a. Use a phasor-diagram analysis to find expressions for the peak currents $I_R$and $I_C$.

b. Complete the phasor analysis by finding an expression for the peak emf current I.

Short answer

(a)The expressions for the peak current $I_R$and $I_C$is role="math" localid="1650014783651" $I_R=\frac{{\epsilon }_o}{R}$and ${\epsilon }^2=(R+\frac{1}{{\omega }^2{C}^2})I^2$.

(b)The expression for peak current is$I=\frac{{\epsilon }_o}{\sqrt{R^2+{X_c}^2}}$

Step-by-step solution

Part (a) Step 1:Given Information

We have to find out the expression for $I_{R}and{I}_C$.

Part (a) Step 2:Formula Used

Now As we know,

$\epsilon ={\epsilon }_o\cos \omega t$and $I=\frac{{\epsilon }_0}{Z}$

Peak current $I_C$is very small and $V_R=I_{R}R$,

From this we get,

role="math" localid="1650015285522" $I_R=\frac{{\epsilon }_o}{R}$

As peak current $I_C$is very small So,$I_R=\frac{{\epsilon }_o}{R}$.

Part (b) Step 1: Given Information

We have to find out the expression for peak emf current I .

Part (b) Step 2: Formula Used

Now as we know,

${{\epsilon }_o}^2={V_R}^2+{V_C}^2$

where role="math" localid="1650015572237" $V_R=IRand{V}_C=I{X}_C$,

putting these values we get,

${{\epsilon }_o}^2={\left(IR\right)}^2+{\left(I{X}_C\right)}^2$

On simplifying we get,

$I=\frac{{\epsilon }_o}{\sqrt{R^2+{X_C}^2}}$,

Part (b) Step 3: Find Values

Now, if we have I

then $V_R=IR=\frac{{\epsilon }_{o}R}{\sqrt{R^2+{\displaystyle \frac{1}{{\omega }^2{C}^2}}}}$

And $V_C=I{X}_C=\frac{{\epsilon }_o/\omega C}{\sqrt{R^2+{\displaystyle \frac{1}{{\omega }^2{C}^2}}}}$

Hence, the peak emf current is$\frac{{\epsilon }_o}{\sqrt{R^2+{\displaystyle \frac{1}{{\omega }^2{C}^2}}}}$