Find the Norton equivalent of the following circuit as seen

Find the Norton equivalent of the following circuit as seen between terminals a - a. In the figure, R_1 =68 k Ohm, R_2 = 27 k Ohm, V_1 = V_2 = 9 V.

Solution

Rth = R1 || R2 = 68*27/(68+27) = 19.33 kOhm

Vth

Assuming Va\' = 0V

Voltage drop across 68kOhm V68 = 68*18/(68+27) = 12.88V

So Vth = Vaa\' = 12.88-9 = 3.88V

So Short circuit current I = Vth/Rth = 0.2 mA

So equivalent norton parameter

Rth = 19.33kOhm

Isc = 0.2mA