Three `200 Omega` resistors are connected as shown in figure. The maximum power that can be dissipated in any one of the resistor is 50 W. Find the total power in watts dissipated in the circuit for maximum voltage across the terminals A and B.

Answer

75

Explanation

The minimum current that can be passed through any resistor is `= sqrt((50)/(200)) = (1)/(2) A`
To keep all resistors safe `I le (1)/(2) A` and also
`(1)/(2) le (1)/(2) rarr I le 1 A`

This means `I lt (1)/(2)A`
`(V_(A)-V_(B))_(max)=(1)/(2)xx300 =150 V`
`P=((150)^(2))/(300)=75 W`.