Question

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 ?

Answer 1

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