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In the circuit shown in figure I(1),I(2)...

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

A

`0.212 MA, 3.32 mA, 3.108 mA`

B

`2.12 mA, 3.32 mA, 3.108 mA`

C

`0.212 mA, 0.332mA, 3.108 mA`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A

`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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