What would be the value of `E` so that the galvanometer `G` circuit shown on deflection

Options

A. `5V`
B. `10V`
C. `15V`
D. `20V`

Answer

B

Explanation

Refer `G` will show no deffection if point `C` and `l` are as the same potential

In circuit ABCDEF current, `I_(1) = (20)/(5+10+5) = 1A`
pot diff between `C` and `D =I_(1) xx 10 = 1xx 10 = 10V`
Since point `D` is a zero potential therefore pot of point `C = 10V`
Hence potential of point `J` potential point `C= 10V`
Since point `M` is at zero potential pot diff between `E and J = (E - 10)V`
In circuit KLMJK current `1` is
`1 = (E - 10)/(5+10+5) = (E - 10)/(20)`
pot diff across `5 Omega` between point `E and J` is
`E - 10 = ((E - 10)/(20)) 5 = (E- 10)/(4)`
or `4E- 40= E- 10 or 3E= 30or E= 10V`