Shown a circuit with known resistance `R,R_(1)` and `R_(2)` neglect the resistance of the conducting wires and internal resistance of current source The magnitude of electrotive force `E_(1)` each that the currect `l` through the resistance `R` in zero will be

Options

A. `E((R_(2))/(R_(1)))`
B. `E((R_(1))/(R_(2)))`
C. `E((R_(1)+R_(2))/(R_(2)))`
D. `E((R_(1))/(R_(1)+R_(2)))`

Answer

c

Explanation

The distribution of current in the various arms as per Kirchhoff's first law, has been

Using Kirchhoff's second law in closed circuit `ABFGCDAZ` and `BFGCB` we have
`E = (I + I_(1)) R_(2) + IR`
and `E_(1) = (I + I_(1)) R_(2) + I_(1)R_(1)`
Putting `I = 0` , we have :
`E= I_(1)R_(2) and E_(1) = I_(1)(R_(1) + R_(2))`
`:. E_(1) = (E )/(R_(2)) (R_(1) + R_(2))`