There was some internal resistance to the revolution - Batteries & Internal Resistance

2.
a) 16 ohm current = .106 mA
b) battery current = 6.37 mA
c) New Req = 1413.8 ohms, terminal voltage = 8.987 V

3.
a) 2 A
b) 8V

From Textbook:
18 a) 8.406 V, b) 8.491 V
20. r = .4068 ohms
21. .06 ohms
29.
For the conservation of current at point c, we have
Iin = Iout ;
I1 = I2 + I3 .
When we add the internal resistance terms for the two
loops indicated on the diagram, we have
loop 1: V1 – I2R2 – I1R1 – I1r1 = 0;
+ 9.0 V – I2(15 W) – I1(22 W) – I1(1.2 W) = 0;
loop 2: V3 + I2R2 + I3r3 = 0;
+ 6.0 V + I2(15 W) + I2(1.2 W) = 0.
When we solve these equations, we get
I1 = 0.60 A, I2 = – 0.33 A, I3 = 0.93 A.

Oh, and here's something interesting I found:

February break clue #2: