If the resistances are chosen for the circuit shown in figure in such a way that no current flows through the battery with emf `E_(1)`, the voltage `V_(2)` across `R_(2)` and the current `I_(3)` flowing through `R_(3)` are respectively,

Options

A. `V_(2)=-4V, l_(3)=5/2 A`
B. `V_(2)=+4V, l_(3)=5/2 A`
C. `V_(2)=-3V, l_(3)=1A`
D. `V_(2)=+3V, l_(3)=2A`

Answer

C

Explanation

Resistance `R_(1)` and `R_(4)` are connected in series, so the circuit can be redrawn as

As given , `I_(1)=0A`, so the potential drop of 2V takes place across the parallel branch AB and resistor `R_(3)`.
Hence, `2=I_(3)R_(3)=I_(3)2`
`rArr I_(3)=1A`
Similarly, `V_(AB)=5-I_(3)R_(2)`
`2=5-1.R_(2)`
`rArr R_(2)=3 Omega`
`rArr V_(2)=-I_(3)R_(2)=-3xx1=-3V`
Hence, the correct option is (c).