Chapter 19, Problem 50P

(a)

To determine

The total capacitance of the circuit.

Answer to Problem 50P

The total capacitance of the circuit is $2.33\times {10}^{-3}\text{F}$ .

Explanation of Solution

Given info:

The total resistance of the circuit is, $R=15\text{kΩ}=15\times {10}^{-3}\Omega$ .

The emf of the battery is, $V=25\text{V}$ .

The time constant of the circuit is, $\tau =35\text{μs}=35\times {10}^{-6}\text{s}$ .

Formula Used:

The expression to calculate the total capacitance of the circuit is,

  $C=\frac{\tau }{R}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}C=\frac{35\times {10}^{-6}\text{s}}{15\times {10}^{-3}\Omega }\\ =2.33\times {10}^{-3}\text{F}\end{array}$

Conclusion:

Thus, the total capacitance of the circuit is $2.33\times {10}^{-3}\text{F}$ .

(b)

To determine

The time to reach at the voltage 16 V across resistor.

Answer to Problem 50P

The time to reach at voltage 16 V across resistor is $38.80\times {10}^{-6}\text{s}$ .

Explanation of Solution

Given info:

The emf of the battery is, $V=24\text{V}$

The voltage across the resistor is, $V_1=16\text{V}$ .

Formula Used:

The expression to calculate the time to reach the voltage across the resistor is,

  $V_1=V\left(1-e^{-\frac{t}{\tau }}\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}16\text{V}=24\text{V}\left(1-e^{-\frac{t}{35\times {10}^{-6}\text{s}}}\right)\\ 1-e^{-\frac{t}{35\times {10}^{-6}\text{s}}}=0.67\\ e^{-\frac{t}{35\times {10}^{-6}\text{s}}}=0.33\\ -\frac{t}{35\times {10}^{-6}\text{s}}=\ln 0.33\\ t=38.80\times {10}^{-6}\text{s}\end{array}$

Conclusion:

Thus, the time to reach at voltage 16 V across resistor is $38.80\times {10}^{-6}\text{s}$ .