Chapter 9, Problem 1PP

To determine

The unknown parameters using Kirchhoff’s Laws.

Answer to Problem 1PP

ES1= 12 V	E1= 0.89 V	E2= 20.89 V	E3= 11.1 V
ES2= 32 V	I1= 1.31 mA	I2= 20.89 mA	I3= 22.2 mA
	R1= 680 Ω	R2=1000 Ω	R3=500 Ω

Explanation of Solution

We will use Kirchhoff’s Voltage law to solve for the unknown parameters.

Consider two loops LOOP 1 and LOOP 2 with their respective loop currents I₁ and I₂

  

Using KVL in loop 1,

  $\begin{array}{l}+12-500\left(I_1+I_2\right)-680I_1=0\\ -\text{1180}{I}_1-500I_2+12=0\mathrm{...}\text{ (1)}\end{array}$

Using KVL in loop 2,

  $\begin{array}{l}+32-500\left(I_2+I_1\right)-1000I_2=0\\ -500I_1-1500I_2+32=0\mathrm{...}\text{ (2)}\end{array}$

From equation (1), we can write I₂as ,

  $I_2=\frac{12-\text{1180}{I}_1}{500}\mathrm{...}\text{ (3)}$

Substitute this expression for I₂in equation (2),

  $\begin{array}{l}-500I_1-1500\left(\frac{12-\text{1180}{I}_1}{500}\right)+32=0\\ -500I_1-3\left(12-\text{1180}{I}_1\right)+32=0\\ 3040I_1=4\\ I_1=\frac{4}{3040}\\ I_1=1.31\text{ mA }\end{array}$

Substitute the value of I₁ in equation (3).

  $\begin{array}{l}{I}_2=\frac{12-\text{1180}\left(1.31\times {10}^{-3}\right)}{500}\\ =\frac{12-\text{1}\text{.55}}{500}\\ =20.89\text{ mA }\end{array}$

CurrentI₃ is the sum of currents I₁ and I₂

  $\begin{array}{l}{I}_3=I_1+I_2\\ =\left(1.31+20.89\right)\text{ mA}\\ =22.2\text{ mA}\end{array}$

The voltage drops are obtained by using Ohm’s Law,

  $\begin{array}{l}{E}_1=I_1{R}_1\\ =\left(1.31\times {10}^{-3}\right)\times \left(680\right)\\ =0.89\text{ V}\end{array}$

  $\begin{array}{l}{E}_2=I_2{R}_2\\ =\left(20.89\times {10}^{-3}\right)\times \left(1000\right)\\ =20.89\text{ V}\end{array}$

  $\begin{array}{l}{E}_3=I_3{R}_3\\ =\left(22.2\times {10}^{-3}\right)\times \left(500\right)\\ =11.1\text{ V}\end{array}$

The results are summarized in the figure below,