Three 100 Ω resistors are connected as shown in the figure. The maximum power that can safely be delivered to any one resistor is 20.5 W.

Question

(a) What is the maximum potential difference that can be applied to the terminals a and b?

(b) For the voltage determined in part (a), what is the power delivered to each resistor?

1. resistor on the left:
2. resistor at the top of the loop:
3. resistor at the bottom of the loop:

(c) What is the total power delivered to the combination of resistors?

Answer 1

Since each resistor has the same resistance R = 100Ω, the current I is equally split and the total potential drop must be V = IR + (I/2)R = 3IR/2. Therefore V_max = 3I_maxR/2. I_max is determined by the left resistor (since the loop resistors get half of I_max), so solve 20.5W = P_max = I_max²R for I_max to get V_max = 3√(P_maxR)/2.