Question

Electric circuits problem?

Answer 1

(a) `v_{o_1} =11.3` V, `v_{o_2} = 0.3` V
(b) `v_{o_1} = -9` V, `v_{o_2} = -6.6` V

Explanation:

A forward biased Silicon diode drops around 0.7 V, while a forward biased Germanium diode drops around 0.3 V.

In (a) both the diodes are forward biased.
`v_{o_1} = 12 \mbox{ V}-0.7 \mbox{ V} = 11.3`V
`v_{o_2} = 0.3` V

In (b) both diodes are again forward biased:
`v_{o_1} = -10\mbox{ V} +0.7\mbox{ V}+0.3\mbox{ V} = -9` V
`v_{o_2} = {3.3 \mbox{ k}}/{3.3\mbox{ k} + 1.2 \mbox{ k}}times(-9 \mbox{ V}) = -6.6` V