16 V Si Si 47 k2 4 V V SolutionThe three diodes are on and e

+16 V Si Si 4.7 k2 -4 V V

The three diodes are on and each diode has drope is 0.7v

so,the current is flow through the circuit,hence consider any path from supply voltage to negative supply voltage..

take a KVL to the circuit path,

-16V+0.7V+0.7+I(4.7Kohm)-4v=0

I=(20-1.4)/4.7kohm

I=18.6/4.7kohm

I=3.9mA

OUTPUT EQUATION

-V_O +3.9mA(4.7k)-4V=0

V_O =18.33-4=14.44V