The ideal diodes `D_(1)` and `D_(2)` are to be connected in the positions shown in the circuit shown below

calculate the current in the circuit
(i) when p side of `D_(1)` is connected to resistance `R_(1)` and p side of `D_(2)` is connected to resistance `R_(2)`
(ii) when n side of `D_(1)` connected to resistance `R_(1)` and n side of `D_(2)` is connected to resistance `R_(2)`

Explanation

(i) The given arrangement is shown below :

In this case `D_(1)` becomes forward d biased and `D^(2)`
becomes forward biased and `D^(2) ` becomes reverse biased So , the circuit can be d simplified as shown under.

Current flowing in the circuit is
`I= (5)/(R_(1)+R_(2)) =(5)/( 20+10) = (5)/( 30) = 0.167A`
(ii) The given arrangement is shown below

In this case `D_(1)` becomes reverse biased and `D_(2)` becomes forward biased So, the circuit can be simplified as shown under.

Current in the circuit will be
`I = (5)/( R_(2)+R_(3)) =(5)/( 30+10) = (5)/( 4) = 0.125 A`