Chapter 21, Problem 71PE

To determine

(a)

The time constant with resistance (in ohms) and capacitance (in farads).

Answer to Problem 71PE

  $20.0\text{ seconds}$.

Explanation of Solution

Given information:

Figure shows how a bleeder resistor is used to discharge a capacitor after an electronic devices is shut off, allowing a person to work on the electronics with less risk of shock.

Calculation:

In RC circuit, the time constant ( $\tau$ ) is the product of resistance (in ohms) and capacitance (in farads) i.e. $\tau =RC$.

  $\begin{array}{l}\text{The given resistance(}R\text{)=250}\times {\text{10}}^3\Omega \text{ and capacitance(}C\text{)=80}\times {\text{10}}^{-6}F\text{ are:}\\ \Rightarrow \tau =RC\text{ {The time constant of }RC\text{ circuit}}\\ \Rightarrow \tau =(\text{250}\times {\text{10}}^3)\times (\text{80}\times {\text{10}}^{-6})\text{ {Substitute, }R\text{=250}\times {\text{10}}^3\Omega \text{ and }C\text{=80}\times {\text{10}}^{-6}F\text{}}\\ \Rightarrow \tau =20.0\text{ seconds}\end{array}$

Therefore, the time constant is $20.0\text{ seconds}$.

To determine

(b)

The reduction of the voltage on the capacitor to one quarter percent of its full value.

Answer to Problem 71PE

  $120\text{ s or (2 minutes)}$

Explanation of Solution

Given information:

Figure shows how a bleeder resistor is used to discharge a capacitor after an electronic devices is shut off, allowing a person to work on the electronics with less risk of shock.

It takes to reduce the voltage on the capacitor to $0.250\%(5\%\text{ of 5\%)}$ of its full value once discharge begins.

Calculations:

To find how long will it take to reduce the voltage on the capacitor to $0.250\%(5\%\text{ of 5\%)}$ of its full value, they are using the formula for the voltage of a discharging capacitor as follows:

  $\begin{array}{l}\Rightarrow \text{To find how long will it take to reduce the voltage on capacitor are:}\\ V=V{e}^{-t/RC}\text{ (i) {Formula for finding the voltage on discharging capacitor}}\\ \text{The voltage falls to 0}\text{.250\% of its initial value then}\\ V=V_0\cdot e^{-1}\text{ (ii)}\\ V=0.0025V_0\text{ {0}\text{.250\%=}\frac{0.250}{1000}\text{=0}\text{.0025}}\\ \text{Substitute the above value in the equation (i)}\\ V=V{e}^{-t/RC}\end{array}$

  $\begin{array}{l}t=-RC\text{ In }\left(\frac{V}{V_0}\right)\\ t=-(\text{250}\times {\text{10}}^3)\cdot (\text{80}\times {\text{10}}^{-6})\text{In }\left(\frac{0.0025V_0}{V_0}\right)\text{ {where,}R\text{=250}\times {\text{10}}^3\Omega \text{ and }C\text{=80}\times {\text{10}}^{-6}F\text{}}\\ t=-20\text{In }0.0025\\ t=119.8292909\\ t=120\text{ s (approx}\text{.) or (2 minutes)}\end{array}$

It takes $120\text{ s or (2 minutes)}$ to reduce the voltage on capacitor.

To determine

(c)

The time takes to rise the initial voltage.

Answer to Problem 71PE

  $16\text{ ms}$.

Explanation of Solution

Given information:

If the capacitor is charged to a voltage $V_0$ through a $100\Omega$ resistance, the time it takes to rise to $0.865V_0$.

Calculation:

To find the time it takes to rise $0.865V_0$ ,by using the formula for the voltage of a discharging capacitor as follows:

  $\begin{array}{l}\Rightarrow \text{The voltage on capacitor is}\\ V=\text{emf (1}-e^{-1})\\ \text{ =emf(1}-0.865\text{)}\\ \text{ =emf(0}\text{.135)}\\ \frac{V}{V_0}=0.135\end{array}$

  $\begin{array}{l} \Rightarrow \text{To find the time it takes to rise to 0}\text{.865 }{V}_0\text{:}\\ V=V{e}^{-t/RC}\text{ {Formula for finding the voltage on discharging capacitor}}\\ t=-RC\text{ In }\left(\frac{V}{V_0}\right)\\ t=-(100)\cdot (\text{80}\times {\text{10}}^{-6})\text{In }\left(0.135\right)\text{ {where,}R\text{=100}\Omega \text{ and }C\text{=80}\times {\text{10}}^{-6}F\text{}}\\ t=-0.008\text{In }0.135\\ t=0.16019844\\ t=16\text{ ms }\end{array}$

It takes $t=16\text{ ms}$ to rise up the initial voltage.