A 10 A source having 100 ohm shunt internal resistance will deliver maximum power at which condition

by

Statement: A 10 A source has 100-ohm shunt internal resistance. The maximum power than can be delivered to the load is:

  1. 100 W
  2. 1000 W
  3. 2500 W
  4. 10000 W

Correct answer: 3. 2500 Watt

Solution:

Maximum power will be delivered when RL = Rs = 100 ohms

According to current divider rule Is = (100 ohm)/(100 ohm+100 ohm) * 10 A= 5 A

Now Ps = (Is)2.Rs = (5)2 A * 100 ohms = 2500 W

Leave a Comment