Circuit Theorems and Conversions - Engineering

Q1:

A 680  load resistor, RL, is connected across a constant current source of 1.2 A. The internal source resistance, RS, is 12 k. The load current, RL, is

A 0 A

B 1.2 A

C 114 mA

D 1.14 A

ANS:D - 1.14 A

RL = 12 X 10^3 ohms.
r=680 ohms.
I=1.2 A.
Solution
I load=1.2((680/(680+12000)),
I=1.14A.