Download Chapter 25 Problem 50 † Solution Find the current through the

Chapter 25
Problem 50
2.0 Ω
†
2.0 Ω
A
6.0 V
2.0 Ω
4.0 Ω
Solution
Find the current through the ammeter.
On the diagram 6 currents are indicated. One current going through each component of the circuit.
I1
F
2.0
I2
B
6.0 V
I4
2.0
A
I3
2.0
I6
C
A
I5
D
4.0
E
Using Kirchoff’s rules we need 6 equations. 3 of the equations can be obtained from the voltage around a
loop. Using the 3 smallest loops that completely cover the circuit gives
6.0 V − 2.0 ΩI2 − 2.0 ΩI4 = 0
−2.0 ΩI2 + 2.0 ΩI3 = 0
−2.0 ΩI4 + 4.0 ΩI5 = 0
(LoopEF ABD)
(LoopABCA)
(LoopBDCB)
Notice the last two loops involve the ammeter. However, there is no voltage drop across a perfect
ammeter and nothing shows up in the equations. The nodes can be taken at points A, D, and B.
I1 = I2 + I3
I4 + I5 = I1
I2 + I6 = I4
(N odeA)
(N odeD)
(LoopBDCB)
Now solve these 6 equations for 6 unknowns. Substitute the Node D equation into the remaining 5
equations to eliminate I1 . The relative order of the equations are maintain and we, therefore, have
6.0 V − 2.0 ΩI2 − 2.0 ΩI4 = 0
−2.0 ΩI2 + 2.0 ΩI3 = 0
−2.0 ΩI4 + 4.0 ΩI5 = 0
I4 + I5 = I2 + I3
I2 + I6 = I4
Use the last equation to eliminate I2 in the other 4 equations
6.0 V = 4.0 ΩI4 + 2.0 ΩI6 = 0
†
Problem from Essential University Physics, Wolfson
−2.0 ΩI4 + 2.0 ΩI6 + 2.0 ΩI3 = 0
−2.0 ΩI4 + 4.0 ΩI5 = 0
I5 = −I6 + I3
Use the last equation to eliminate I3 in the other 3 equations
6.0 V − 4.0 ΩI4 + 2.0 ΩI6 = 0
−2.0 ΩI4 + 4.0 ΩI6 + 2.0 ΩI5 = 0
−2.0 ΩI4 + 4.0 ΩI5 = 0
Use the last equation to eliminate I4 in the other 2 equations
6.0 V − 8.0 ΩI5 + 2.0 ΩI6 = 0
4.0 ΩI6 − 2.0 ΩI5 = 0
Use the last equation to eliminate I5 in the other equation
6.0 V − 14.0 ΩI6 = 0
Therefore, I6 is
I6 =
6.0 V
= 0.429 A
14.0 Ω