The 5.00-V battery in (Figure 1) is removed from the circuit and replaced by a 1

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.

I=

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.

I=

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.

I=

Part D

Find the potential difference Vab of point a relative to point b.

Express your answer with the appropriate units.

I=

Part D

Find the potential difference Vab of point a relative to point b.

Express your answer with the appropriate units.

Explanation / Answer

let,

current in upper branch is i1

current in middle branch is i2

current in lower branch is i3

by applying Kirchhoff's voltage rule for upper loop,

10-i1*2-i2*1-15-i2*4-i1*3=0 ---(1)

by applying Kirchhoff's voltage rule for lower loop,

15+i2*1-i3*10+i2*4=0 -----(2)

by applying Kirchhoff's current rule for at junction,

i1=i2+i3 ----(3)

from equation no (1), (2) and (3)

i1=0

i2=1A

i3=1A

A)

current in the upper branch i1=0

B)

current in middle branch i2=1A (towards left)

C)

current in lower branch i3=1A (towards right)

D)

potentil difference is,

Vab=i2*4 -i1*3

Vab=1*4-0

Vab=4 V

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