1.

In the circuit shown in figure I_(1),I_(2) and I_(D_(2)) are respectively-

Answer»

`0.212 MA, 3.32 mA, 3.108 mA`
`2.12 mA, 3.32 mA, 3.108 mA`
`0.212 mA, 0.332mA, 3.108 mA`
NONE of these

Solution :`I_(1) = (V_(1))/(R_(1)) = 0.212 mA`
By KVL,
`- V_(T_(1)) - V_(T_(2)) - V_(2)+E = 0`

`:. V_(2) = 20 - 0.7 - 0.7 = 18.6 V`
`I_(2) = (V_(2))/(R_(2)) = 3.32 mA`


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