Question

Pls answer in detail.... Points will be awarded

Answer 1

rearrange the circuit , new arrangement is shown in figure.

now here it is clear that two resistors 6Ω and 3Ω are joined in parallel.

so, R1 = 6 × 3/(6 + 3) = 2Ω

R1 and 4Ω are joined in series so, R2 = R1 + 4Ω = 2Ω + 4Ω = 6Ω

now R2 and 3Ω are joined in parallel combination. so, Req = R2 × 3/(R2 + 3)

= 6 × 3/(6 + 3) = 2Ω

applying Ohm's law,

V = IR

so, I = V/R = 6 V/2Ω = 3A

hence, current through circuit should be 3A.