In the circuit shown, the cell is ideal with emf = `2V` . The resistance of the coil of the galvanometer `G` is `1 Omega`.

(i) No current flows in `G`
(ii) `0.2A` current flows in `G`
(iii) potential difference across `C_(1) is 1V`
(iv) Potential difference across `C_(2)` is `1.2V`

Options

A. (i),(iii)
B. (i),(ii),(iv)
C. (ii),(iii),(iv)
D. (i),(ii),(iv)

Answer

C

Explanation

`i=(2)/(4+1+5) = 1/5A = 0.2 A`
p.d. across `C_(1)`:
`V_(A) - ixx4-ixx1=V_(C)`
`V_(A)-V_(B)=5i=1V`
p.d across `C_(2)`:
`V_(B) - ixx1-ixx5=V_(D)`
`V_(D)-V_(D)=6i=1.2V`.