The 5.00-V battery in (Figure 1) is removed from the circuit and replaced by a 1
ID: 1541081 • Letter: T
Question
The 5.00-V battery in (Figure 1) is removed from the circuit and replaced by a 15.00-V battery, with its negative terminal next to point b. The rest of the circuit is as shown in the figure.
Part A
Find the current in the upper branch.
Express your answer with the appropriate units. Enter positive value if the current is to the left and negative value if the the current is to the right.
Part B
Find the current in the middle branch.
Express your answer with the appropriate units. Enter positive value if the current is to the left and negative value if the the current is to the right.
Part C
Find the current in the lower branch.
Express your answer with the appropriate units. Enter positive value if the current is to the left and negative value if the the current is to the right.
2.000 10.00 V +I, a 3.00 Q 1.00 n 5.00V 4.00 Q 10.00 (2 AWWAExplanation / Answer
Use Kirchhoff’s rules
since the 15 V battery has the largest voltage assume I1 is to the right through the 10V battery I2 is to the left through the 20V battery and I3 is to the right through the 10 ohm resistor
in each loop counterclockwise direction
in upper loop
10V + (2+3)I1 + (1+4)I2 -15V = 0
-5V + 5I1 + 5I2 = 0
I1+I2 = 1 ...(1)
lower loop
15 - 5I2-10*I3 = 0
I2-2*I3 = 3
using junction law = I2 = I1+I3
I2 - 2*(I2-I1) = 3
I2 - 2I2 + 2I1 = 3
2I1 - I2 = 3 .....(2)
add eq(1) and (2)
3I1 = 4
I1 = 4/3
I2 = -1/3
I3 = -5/3
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