A 2.5-kohms and a 3.7-kohms resistor are connected in parallel; this combination is connected in series with a 1.4 kohms resistor. If each resistor is rated at 0.5W (maximum without overheating), what is the maximum voltage that can be applied across the whole network?

Answer 1

R1 = 1.4k Ohms
R2 = 2.5k Ohms
R3 = 3.7k Ohms

P1 max = I1^2*R1 = 500 mW
I1^2*1.4 = 500
I1^2 = 357.14
I1 = 18.9mA, max = Total current.

Req. = R1 + (R2*R3)/(R2+R3) =
1.4 + (2.5*3.7)/(2.5+3.7) = 1.4 + 1.49 =
2.89k Ohms

E max = I1*Req = 18.9 * 2.89 = 54.6 V. =