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Consider the circuit shown in the figure...

Consider the circuit shown in the figure. The current `I_(3)` is equal to

A

5amp

B

2amp

C

`-3amp`

D

`-5//6amp`

Text Solution

Verified by Experts

The correct Answer is:
D
After applying KVL for loop (1) and loop (2)
We get 28i,=-6-8 `rArr` `i_(1)`=`-(1)/(2)A`
and `54i_(2)` =-6-12 `rArr` `i_(2)`=`-(1)/(3)A`
Hence `i_(3)`=`i_(1)_i_(2)`=-(5)/(6)A`
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