Question

Electric circuits?

Answer 1

(a) `V_{o} = 19.3` V, `I_D = 2.05` mA
(b) `V_{o} = 14.3` V, `I_D = 6.5` mA

Explanation:

In (a) both the forward biased Silicon diodes drop 0.7 V. Thus
`V_{o} = 20` V - 0.7 V = 19.3 V
The current through the 4.7 k resistance is
`{19.3 \mbox{ V}}/{4.7 \mbox{ k}} ~~ 4.11` mA

This current is distributed equally through the two forward biased diodes.
`I_D = {4.11\mbox{ mA}}/2=2.05` mA

In (b), only one of the two parallel diodes is forward biased.
`V_o = 15`V - 0.7 V = 14.3 V
The current `I_D` in this case is the same as that through the 2.2 k resistor.
`I_D = {14.3 \mbox{ V}}/{2.2 \mbox{ k}}` = 6.5 mA