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Question

In the circuit shown in figure, find the current through the branch BD
119067.png

A
5A
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B
4A
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C
3A
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D
2A
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Solution

The correct option is A 5A
From figure:-

In Loop ABDA:-

V1=R1i2+R3(i1+i2)

3i2+6i2=15

i1+3i2=5

i1=53i2 -------(a)

In Loop BCDB:-

V2=R2i1+R3(i1+i2)

6i1+3i2=30

2i1+i2=10

2(53i2)+i2=10 from equation (a)

105i2=10

i2=0

from equation (a):-

i1=53i2=5A

Current in branch BD=i1+i2=5A

Answer-(A)

837266_119067_ans_a104f993bc6a43c89883c28a68b10750.jpg

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