Chapter 19, Problem 52P

(a)

To determine

The potential at the point a .

Answer to Problem 52P

The potential at the point a is $8\text{V}$ .

Explanation of Solution

Given info:

The resistance of the first resistor is, $R_1=8.8\Omega$ .

The resistance of the second resistor is, $R_2=4.4\Omega$ .

The capacitance of the first capacitor is, $C_1=0.48\text{μF}=4.8\times {10}^{-7}\text{F}$ .

The capacitance of the second capacitor is, $C_2=0.24\text{μF}=2.4\times {10}^{-7}\text{F}$ .

The potential difference of the battery is, $V=24\text{V}$ .

Formula Used:

The expression to calculate the potential at the point a is,

  $V_a=V\left(\frac{R_2}{R_1+R_2}\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{V}_a=\left(24\text{V}\right)\left(\frac{4.4\Omega }{8.8\Omega +4.4\Omega }\right)\\ =8\text{V}\end{array}$

Conclusion:

Thus, the potential at the point a is $8\text{V}$ .

(b)

To determine

The potential at the point b.

Answer to Problem 52P

The potential at the point b is $8\text{V}$ .

Explanation of Solution

Given info:

The capacitance of the first capacitor is, $C_1=0.48\text{μF}=4.8\times {10}^{-7}\text{F}$ .

The capacitance of the second capacitor is, $C_2=0.24\text{μF}=2.4\times {10}^{-7}\text{F}$ .

Formula Used:

The expression to calculate the potential at the point b is,

  $V_b=V\left(\frac{C_2}{C_1+C_2}\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{V}_b=\left(24\text{V}\right)\left(\frac{2.4\times {10}^{-7}\text{F}}{4.8\times {10}^{-7}\text{F}+2.4\times {10}^{-7}\text{F}}\right)\\ =8\text{V}\end{array}$

Conclusion:

Thus, the potential at the point b is $8\text{V}$ .

(c)

To determine

The final potential at the point b

Answer to Problem 52P

The final voltage of the point b is 8 V.

Explanation of Solution

Introduction:

The charge in the each capacitor remains same for the series combination of the capacitors.

If the switch is closed in the circuit then uncharged capacitors become in series combination. The charge in both the capacitor becomes same due to series combination. The potential at point a and point b is same, so no current flows across both capacitors and point b remain at same voltage that is 8 V.

Conclusion:

Thus, the final voltage of the point b is 8 V.

(d)

To determine

The charge flow through the closed switch.

Answer to Problem 52P

The charge flow through the closed switch is $-57.6\times {10}^{-7}\text{C}$ .

Explanation of Solution

Formula Used:

The expression to calculate the charge flow through the closed switch is,

  $Q=C_2{V}_b-C_1\left(V-V_b\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}Q=\left(2.4\times {10}^{-7}\text{F}\right)\left(8\text{V}\right)-\left(4.8\times {10}^{-7}\text{F}\right)\left(24\text{V}-8\text{V}\right)\\ =-57.6\times {10}^{-7}\text{C}\end{array}$

Conclusion:

Thus, the charge flow through the closed switch is $-57.6\times {10}^{-7}\text{C}$ .