Chapter 19, Problem 79GP

(a)

To determine

The current flowing through the 14-volt battery.

Answer to Problem 79GP

The current flowing through the 14-volt battery us $6.7\times {10}^{-5}\text{A}$ .

Explanation of Solution

Given info:

Consider the given circuit.

  

Formula Used:

Apply Kirchhoff’s in the loop acd.

  $-14+\left(10\times {10}^3\right)I_1+\left(12\times {10}^3\right)I_4=0$    … (1)

Apply Kirchhoff’s in the loop bce.

  $-18+\left(15\times {10}^3\right)I_2+\left(20\times {10}^3\right)I_5=0$     … (2)

Apply Kirchhoff’s in the loop cde.

  $-12-\left(20\times {10}^3\right)I_5+\left(12\times {10}^3\right)I_4=0$     … (3)

Apply Kirchhoff’s current law at node d.

  $I_4=I_1+I_3$    … (4)

Apply Kirchhoff’s current law at node e.

  $I_2=I_3+I_5$     … (5)

Calculation:

Substitute the values from equation (4) and (5) to equation (3).

  $\begin{array}{c}-12-\left(20\times {10}^3\right)\left(I_2-I_3\right)+\left(12\times {10}^3\right)\left(I_1+I_3\right)=0\\ 12I_1+12I_3-20I_2-20I_3=12\times {10}^{-3}\\ 12I_1-20I_2-8I_3=12\times {10}^{-3}\end{array}$     … (6)

Substitute the values from equation (5) to equation (2).

  $\begin{array}{c}-18+\left(15\times {10}^3\right)I_2+\left(20\times {10}^3\right)\left(I_2-I_3\right)=0\\ 15I_2+20I_2-20I_3=18\times {10}^{-3}\\ 35I_2-20I_3=18\times {10}^{-3}\end{array}$    … (7)

Substitute the values from equation (4) to equation (1).

  $\begin{array}{c}-14+\left(10\times {10}^3\right)I_1+\left(12\times {10}^3\right)\left(I_1+I_3\right)=0\\ 10I_1+12I_1+12I_3=14\times {10}^{-3}\\ 22I_1+12I_3=14\times {10}^{-3}\end{array}$    … (8)

On solving equation (6), (7) and (8),

  $\begin{array}{l}{I}_1=6.7\times {10}^{-5}\text{A}\\ I_2=\\ I_3=1.044\times {10}^{-3}\text{A}\end{array}$

Hence, the current flowing through the 14-volt battery us $6.7\times {10}^{-5}\text{A}$ .

Conclusion:

Therefore, the current flowing through the 14-volt battery us $6.7\times {10}^{-5}\text{A}$ .

(b)

To determine

The potential difference between point a and b.

Answer to Problem 79GP

The difference in the voltage is $-16\text{V}$ .

Explanation of Solution

Given information:

Consider the given circuit.

  

Formula Used:

Let $V_{\text{a}}$ is the voltage at node (a) and $V_{\text{b}}$ is the voltage at node (b).

  $\begin{array}{l}{V}_b=I_b{R}_2+I_1{R}_1=V_a\\ V_a-V_b=-I_2{R}_2+I_1{R}_1\end{array}$

Calculation:

The values of $I_5$ is calculated as,

  $\begin{array}{c}{I}_5=\frac{18-15000\left(1.044\times {10}^{-3}\text{A}\right)}{35000}\\ =6.68\times {10}^{-5}\text{A}\end{array}$

Then, the value of $I_2$ is obtained as,

  $I_2=6.68\times {10}^{-5}+1.044\times {10}^{-3}\text{A}$

The difference between the voltage is calculated as,

  $\begin{array}{c}{V}_a-V_b=-\left(1.11\times {10}^{-3}\right)\left(15000\right)+\left(6.7\times {10}^{-5}\right)\left(10000\right)\\ =-16\text{V}\end{array}$

Conclusion:

Therefore, the difference in the voltage is $-16\text{V}$ .