The resistance R1 is connected in a circuit, and the consumed electric power is P1. The other resistance R2 is connected in series with it. The electric power of R2 is P2, P1: P2 = 36:5, and R1 > R If resistance R1 is connected in a circuit, the consumed electric power is P1, and another resistance R2 is connected in series with it in the same circuit, the consumed electric power of R2 is P2, P1: P2 = 36:5, and R1 > R2, then R1: R2 =?

The resistance R1 is connected in a circuit, and the consumed electric power is P1. The other resistance R2 is connected in series with it. The electric power of R2 is P2, P1: P2 = 36:5, and R1 > R If resistance R1 is connected in a circuit, the consumed electric power is P1, and another resistance R2 is connected in series with it in the same circuit, the consumed electric power of R2 is P2, P1: P2 = 36:5, and R1 > R2, then R1: R2 =?

According to P = I ^ 2R,
It can be calculated by formula
36=I^2R1,5=I^2R2
therefore
R1=36/I^2
R2=5/I^2
So R1: R2 = 36:5