Chapter 31: Q. 41 (page 903)

A $1.0 μ F$ capacitor is discharged, starting at $t = 0$ . The displacement current through the plates is $I_{d i s p} = (10 A) e x p (- t / 2.0 μ s)$ . What was the capacitor’s initial voltage ${(∆ V_{c})}_{o}$ ?

Short Answer

Expert verified

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

Step by step solution

: Given information

We have given that

$C =$ $1.0 μ F$ at $t = 0$

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

Simplification

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

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

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

$\int d Q = \int I_{d i s p} d t$ integrate both side.

$Q = \int (10) e^{(- t / 2 μ s)} d t$ = $\frac{(10) e^{(- t / 2 μ s)}}{- 1 / 2 μ s}$

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

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

$∆ Q = 20 (10^{- 6}) C$

$∆ Q = 20 μ C$

We know that

$∆ Q = C ∆ V$

put value of $∆ Q$ $= 20 μ C$ and C $= 1.0 μ F$

$20 μ C = 1 μ F (∆ V)$

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

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A $1.0 μ F$ capacitor is discharged, starting at $t = 0$ . The displacement current through the plates is $I_{d i s p} = (10 A) e x p (- t / 2.0 μ s)$ . What was the capacitor’s initial voltage ${(∆ V_{c})}_{o}$ ?

Short Answer

Step by step solution

: Given information

Simplification

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A 1.0μF capacitor is discharged, starting at t=0. The displacement current through the plates is Idisp=(10A)exp(-t/2.0μs). What was the capacitor’s initial voltage (∆Vc)o?

Short Answer

Step by step solution

: Given information

Simplification

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Physical Quantities And Units

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Study anywhere. Anytime. Across all devices.

A $1.0 μ F$ capacitor is discharged, starting at $t = 0$ . The displacement current through the plates is $I_{d i s p} = (10 A) e x p (- t / 2.0 μ s)$ . What was the capacitor’s initial voltage ${(∆ V_{c})}_{o}$ ?