In the circuit shown the reading of voltmeter `V_(3)` and Ammeter 'A' will be

Options

A. 1.43A
B. 3.13 A
C. 2.5 A
D. 10.0 A

Answer

C

Explanation

2.5A
Since the `D_(1)` diode is in reversed bias, so current will not flow hence no current flow from `R_(2)` hence neglecting `R_(2)`, current flow `R_(1)` from battery, `D_(2)` and `R_(3)`.
`therefore` Current `I=(epsilon)/(R_(1)+D_(2)+R_(3))`
`=(10)/(2+0+2)`
`=(10)/(4)=2.5A`
[Diode is an ideal so its resistance is zero]