Question

# At what rate is electric energy converted to internal energy in the resistors R1 and R2...

At what rate is electric energy converted to internal energy in the resistors R1 and R2 in the figure? Let V1 = 9.60 V, V2 = 1.20 V, R1 = 3.60 ?, R2 = 3.40 ?, and R3 = 7.60 ?

Kirchoff's 2nd Law requires that the algebraic sum of potential differences around each loop be 0.
For the upper loop, then, I1 R1 - I2 R2 must be 7.5 V.
For the inner loop, we have I2 R2 + I3 R3 = 1.6 V.
From Kirchoff's 1st Law, we have I1 + I2 = I3.
We now have 3 equations in the 3 unknown currents:
(A) 3.1 I1 - 3.0 I2 = 7.5
(B) 3.0 I2 + 8.4 I3 = 1.6
(C) I1 + I2 - I3 = 0

Eliminate I1 from A and C:
(D) 6.1 I2 - 3.1 I3 = -7.5
(3.1xB) 9.3 I2 + 26.04 I3 = 4.96
(8.4xD) 51.24 I2 - 26.04 I3 = -63.00
60.54 I2 = -58.04
I2 = -0.9857 amps...and this current runs BACKWARDS through the smaller battery!
I3 = (1.6 + 2.876)/8.4 = 0.5329 amps
I1 = 1.5186 amps, runs in the "right" direction.

"Electric energy converted to internal energy"...P = I^2 R.
In R1, P = about 7 watts, but use calculator
In R2, P = about 3 watts, but use calculator
Total P for these two resistors is about 10 watts

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