Considering the circuit and data given in the diagram, calculate the currents flowing in the diodes `D_(1)` and `D_(2)` Forward resistance of `D_(1)` and `D_(2)` is `20 Omega`.
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Explanation

Since the positive terminal of battery is connected to P-type of both diodes `D_(1)` and `D_(2)` , they are forward biased. These diodes are replaced by with their forward resistance as shown is Fig :

the resistance of `20 Omega` and `20 Omega` in parallel,
`(1)/( R)=(1)/(20) +(1)/(20)` (or) `R = (20)/(2) = 10 Omega`
Therefore, total current `I` in the circuit
`I = (1)/(100 + 10) = (1)/(110) amp`
and `I_(1) = I_(2) = (1)/(2) xx (1)/(110) =(1)/(220) amp`.