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Chapter 31: Q. 41 (page 903)

A $1.0 渭 F$ capacitor is discharged, starting at $t = 0$ . The displacement current through the plates is $I_{d i s p} = (10 A) e x p (- t / 2.0 渭 s)$ . What was the capacitor鈥檚 initial voltage ${(鈭� V_{c})}_{o}$ ?

Short Answer

Expert verified

The initial voltage of capacitor is ${(鈭� V_{c})}_{o} = 20 V$

Step by step solution

01

: Given information

We have given that

$C =$ $1.0 渭 F$ at $t = 0$

$I_{d i s p} = (10 A) e x p (- t / 2.0 渭 s)$

02

Simplification

As we know that displacement current $I_{d i s p}$ flow in capacitor.

so, $\frac{d Q}{d t} = I_{d i s p}$

$d Q = I_{d i s p} d t$

$鈭� d Q = 鈭� I_{d i s p} d t$ integrate both side.

$Q = 鈭� (10) e^{(- t / 2 渭 s)} d t$ = $\frac{(10) e^{(- t / 2 渭 s)}}{- 1 / 2 渭 s}$

$Q_{(t = 0) = \frac{(10) e^{(- 0 / 2 渭 s)}}{- 1 / 2 渭 s}}$

$Q_{(t = 0)} = - 20 (10^{- 6}) C$ since $渭 = 10^{- 6}$

$鈭� Q = 20 (10^{- 6}) C$

$鈭� Q = 20 渭 C$

We know that

$鈭� Q = C 鈭� V$

put value of $鈭� Q$ $= 20 渭 C$ and C $= 1.0 渭 F$

$20 渭 C = 1 渭 F (鈭� V)$

${(V_{c})}_{o} = 20 V$

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