Four resistence of `10 Omega, 60Omega,100Omega`and `200Omega`, respectively taken in order are used to form a Wheatstone's bridge . A 15 V battery is connected to the ends of a `200Omega` resistance, the current through it will be

Options

A. `7.5xx10^(-5)A`
B. `7.5xx10^(-4)A`
C. `7.5xx10^(-3)A`
D. `7.5xx10^(-2)A`

Answer

D

Explanation

Here, the resistance to `10 Omega, 60 Omega` and `100 Omega` are in series and they together are in parallel to `200 Omega` resistance, when a potential difference of 15 V is applied across `200 Omega`, then current through it is,
`l=(15)/(200)=7.5xx10^(-2)A`