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In the circuit shown below E-(1) = 4.0V,...

In the circuit shown below `E-(1) = 4.0V, R_(1) = 2Omega, E_(2) = 6.0V, R_(2)= 4 Omega` and `R-(3) = 2Omega`. The current `I_(1)` is

A

1.6A

B

1.8A

C

2.25A

D

1A

Text Solution

Verified by Experts

The correct Answer is:
B
For "loop" `(1) -2i_(1) -2(i_(1) - i_(2))+4`=0`rArr` `2i_(1) –i_(2)` = 2
For "loop"`(2) -4i_(2) -2(i_(1) - i_(2))+6`=0`rArr` `3i_(1) –i_(2)` = 3
After "solving" equation (1) and (11) we get `i_(1)` = 1.8 A and `i_(2)` = 1.6 A.
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