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In the circuit shown in the figure . ...

In the circuit shown in the figure
.
The value of current `I_1` is

A

2A

B

zero

C

2.5 A

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
b. `x/10 + (x+10)/5 + (x-y-20)/5 = 0`
`y/5 + (y+10)/5 + (y+20-x)/5 = 0`
On solving for x and y we get x = 0,
`y = -10`. Hence `I_1 = 0 and I_3 = 2A`, i.e., `I_4` will also be 2A.
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