Two resistors, R1 = 12 Ω, R2 = 24 Ω, are connected in series to a 48V power supply. How much power is consumed by R1? (2) How much heat does R1 and R2 produce per minute?

Two resistors, R1 = 12 Ω, R2 = 24 Ω, are connected in series to a 48V power supply. How much power is consumed by R1? (2) How much heat does R1 and R2 produce per minute?

(1) I = ur1 + R2 = 4812 + 24a = 43a, the electric power consumed by R1 is P1 = i2r1 = (43) 2 × 12W = 21.3w. Answer: the electric power consumed by R1 is 21.3w. (2) Q1 = i2r1t = (43) 2 × 12 × 60j = 1280j, Q2 = i2r2t = (43) 2 × 24 × 60j = 2560j. Answer: the heat generated by R1 and R2 per minute is 1280j and 2560j

As shown in the figure, if three identical resistors R1 = R2 = R3 are connected in the circuit, the ratio of their voltage to the consumed electric power is ()
Sorry, because there is no diagram, R1 and R2 are connected in parallel and then connected in series with R3

If R1 and R2 are connected in series, then R3 is connected in parallel
Then U1: U2: U3 = 1:1:2
P1:P2:P3=1:1:4
If R1 and R2 are connected in parallel, then R3 is connected in series
Then U1: U2: U3 = 1:1:2
P1:P2:P3=1:1:4
If it's a full union or a full string, it's all 1:1:1

Two resistors are connected in parallel, R1 = 10 Ω, R2 = 20 Ω, and the power supply voltage is 6 v. what are the electric power of R1 and R2 respectively

It can be seen that internal resistance is not included
that
I(R1)=U/R1=6/10=0.6
P(R1)=I(R1)^2*R2=0.36*10=3.6W
I(R2)=U/R1=6/20=0.3
P(R2)=I(R2)^2*R2=0.09*20=1.8W

Resistor R1 = 6 Ω and R2 = 12 Ω are connected in parallel in the circuit. The electric power consumed by R1 is measured to be 6 watts. How much is the voltage at both ends of R1?
(2) What is the total power consumed by R1 and R2?

U ^ 2 / R1 = 6, so u = 6V (the voltage at both ends of RI is 6V) P2 = u ^ 2 / r2 = 36 / 12 = 3W, so (what is the total power consumed by R1 and R2) ptotal = P1 + P2 = 6 + 3 = 9W