Question

The current ix in the circuit given in milliampere is

Answer 1

Solution:

By KCL I1 KCL I1 = 10 mA - Ix

By applying KVL in first loop 

= - 1 - 100 I1 + 100 Ix = 0

= -1 - 1 + 100 Ix + 100 Ix = 0

= 200 Ix = 2

Ix = 2200=10 mA