Download Chapter 25 Problem Ex † Solution Find the current in the 2.0Ω

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Transcript 
Chapter 25
1.0
Problem Ex
†
I1
4.0
12 V
I3
6.0 V
3.0 V
2.0
I2
Solution
Find the current in the 2.0Ω resistor, I2 .
From Kirchoff’s rules we can get two voltage loop equations. The first is the loop on the left hand side.
Starting at the 12 V source and going clockwise gives
12 V − 1.0 ΩI1 − 6.0 V + 2.0 ΩI2 = 0
or
6.0 V − 1.0 ΩI1 + 2.0 ΩI2 = 0
(1)
Using the loop on the right side and going clockwise from the 6.0 V source we get
6.0 V − 4.0 ΩI3 + 3.0 V − 2.0 ΩI2 = 0
or
9.0 V − 4.0 ΩI3 − 2.0 ΩI2 = 0
(2)
Using the node in the middle of the diagram gives
I1 + I2 = I3
(3)
Solving for I1 in equation (1) gives
I1 =
6.0 V + 2.0 ΩI2
= 6.0 A + 2.0I2
1.0 Ω
(4)
Solving for I3 in equation (2) gives
I3 =
9.0 V − 2.0 ΩI2
= 2.25 A − 0.5I2
4.0 Ω
(5)
Substituting equation (4) and (5) into (3) gives
6.0 A + 2.0I2 + I2 = 2.25 A − 0.5I2
3.5I2 = −3.75 A
I2 = −1.07 A
Since the current is negative, the current is travelling opposite to the direction indicated in the diagram.
†
Problem from Essential University Physics, Wolfson
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