Chapter 9, Problem 3PP

To determine

The unknown parameters using Kirchhoff’s Laws.

Answer to Problem 3PP

ES1=6 V	E1= 0.89 V	E2= 20.89 V	E3= 11.1 V
ES2=60 V	I1= 1.31 mA	I2= 20.89 mA	I3= 22.2 mA
	R1=1.6kΩ	R2= 1.2 kΩ	R3= 2.4 kΩ

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,

  $\begin{array}{l}+6-2400\left(I_1+I_2\right)-1600I_1=0\\ -4000I_1-2400I_2+6=0\mathrm{...}\text{ (1)}\end{array}$

Using KVL in loop 2,

  $\begin{array}{l}+60-2400\left(I_2+I_1\right)-1200I_2=0\\ -2400I_1-3600I_2+60=0\mathrm{...}\text{ (2)}\end{array}$

From equation (1), we can write I₂as ,

  $I_2=\frac{6-4000I_1}{2400}\mathrm{...}\text{ (3)}$

Substitute this expression for I₂in equation (2),

  $\begin{array}{l}-2400I_1-3600I_2+60=0\\ -2400I_1-3600\left(\frac{6-4000I_1}{2400}\right)+60=0\\ -2400I_1-9+6000I_1+60=0\\ \text{ 360}0I_1+51=0\\ I_1=-\frac{51}{3600}\\ I_1=-14.16\text{ mA }\\ \end{array}$

The negative sign indicates that the direction of current is different from that assumed at the start.

Substitute the value of I₁ in equation (3).

  $\begin{array}{l}{I}_2=\frac{6-4000I_1}{2400}\\ =\frac{6-4000\left(-14.16\times {10}^{-3}\right)}{2400}\\ =26.11\text{ mA }\end{array}$

CurrentI₃ is the sum of currents I₁ and I₂

  $\begin{array}{l}{I}_3=I_1+I_2\\ =\left(-14.16+26.11\right)\text{ mA}\\ =11.95\text{ mA}\end{array}$

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

  $\begin{array}{l}{E}_1=I_1{R}_1\\ =\left(-14.16\times {10}^{-3}\right)\times \left(1600\right)\\ =-22.66\text{ V}\end{array}$

  $\begin{array}{l}{E}_2=I_2{R}_2\\ =\left(26.11\times {10}^{-3}\right)\times \left(1200\right)\\ =31.33\text{ V}\end{array}$

  $\begin{array}{l}{E}_3=I_3{R}_3\\ =\left(11.95\times {10}^{-3}\right)\times \left(2400\right)\\ =28.68\text{ V}\end{array}$

The results are summarized in the figure below,