Three resistances `2 Omega, 3 Omega` and `4 Omega` are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be

Options

A. `5:4:3`
B. `6:3:2`
C. `4:3:2`
D. `6:4:3`

Answer

D

Explanation

We know that potential drop across a resistance,
V = l R
V = constant
i.e., `l prop (1)/(R )`
`l_(1):l_(2):l_(3)=(1)/(R_(1)):(1)/(R_(2)):(1)/(R_(3))=(1)/(2):(1)/(3):(1)/(4)=6:4:3`
As, `V prop l`
`therefore " " V_(1):V_(2):V_(3)=6:4:3`