Chapter 7.4, Problem 43E

a.

To determine

To calculate: The differential equation for $V$ .

Answer to Problem 43E

The differential equation for $V$

  $V=V_0{e}^{\frac{-t}{40}}$

Explanation of Solution

Given information: Suppose that electricity is draining from a capacitor at a rate proportional to the voltage $V$ across its terminals and that, if t is measured in seconds

  $\frac{dv}{dt}=\frac{V}{-40}$

Calculation:

electricity is draining from a capacitor at a rate

  $\begin{array}{l}\frac{dv}{dt}=\frac{V}{-40}\\ \frac{dv}{v}=\frac{dt}{-40}\\ \ln V=\frac{-t}{40}+C\\ \text{at }t=0\text{ ,}V=V_0\\ V=V_0{e}^{\frac{-t}{40}}\end{array}$

b.

To determine

To calculate: Time taken the voltage to drop to $10\%$ of its original value.

Answer to Problem 43E

Time taken the voltage to drop to $10\%$ of its original value

  $t=92.1\text{ seconds}$ .

Explanation of Solution

Given information: The differential equation for $V$

  $V=V_0{e}^{\frac{-t}{40}}$ .

Calculation:

The differential equation for $V$

  $V=V_0{e}^{\frac{-t}{40}}$

Voltage V is $10\%$ of its original value, $V_0$

  $\begin{array}{l}\frac{dv}{dt}=\frac{V}{-40}\\ \frac{dv}{v}=\frac{dt}{-40}\\ \ln V=\frac{-t}{40}+C\\ \text{at }t=0\text{ ,}V=V_{0}10\%\\ V=V_0{e}^{\frac{-t}{40}}\\ V=0.1V_0\\ 0.1V_0=V_0{e}^{\frac{-t}{40}}\\ e^{\frac{-t}{40}}=0.1\\ \frac{-t}{40}=\ln 0.1\\ t=92.1\text{ seconds}\end{array}$