Concept explainers
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,
Using KVL in loop 2,
From equation (1), we can write I2as ,
Substitute this expression for I2in equation (2),
Substitute the value of I1 in equation (3).
CurrentI3 is the sum of currents I1 and I2
The voltage drops are obtained by using Ohm’s Law,
The results are summarized in the figure below,
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Chapter 9 Solutions
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