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In the circuit shown the currents in 4 O...

In the circuit shown the currents in `4 Omega` and `8 Omega` resistances are

A

`2.5 A, 0A`

B

`5 A, 0A`

C

`2.5 A, (40)/(3) A`

D

`5 A, (80)/(3) A`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) Since battery is in parallel with `4 Omega` resistance, it carries current
`I = (10)/(4) = 2.5 A`
In open loop `a_(1) + 10 - 10 = 0 implies V_(a_(1)) = 0 V`
Thus, all resistance connected between `a_(1)` and `g`, including `8 Omega` resistance, carrying zero current.
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