Lesse... as I had important things to do I thought I’d spell it out more in detail.
If you have a 10V (U=10 V) battery capable of providing 10 mAh, it will give you 1 mA (I = 1/1000 A) for 10 hours. This will happen if you short it with a resistance of U/I = 10 / (1/1000) ohms = 10 kohms. The power developed will be P = U * I = 10 * 1/1000 W = 10 mW.
Now add another battery. If you disconnect the +ve lead from bat 1 and connect it to the +ve side of bat 2, then add another lead from the +ve side of bat 1 to the -ve side of bat 2, you have them vired in series.
|--------------[ 10 kohm ] -------------|
|----[ - bat 1 + ] ----- [ - bat 2 + ] ----|
You’ll get U = 20 V over the resistance, with twice the current through the resistance and the batteries (I = 20/10 000 A = 2 mA), meaning you’ll use your batteries at twice the rate, giving you (10/1000)/(2/1000) = 5 hours of use. The power turned into heat in the resistor will be P = U * I = 20*2/1000 = 40 mW. Each battery will provide 20 mW of power (P=10 * 2/1000 W).
Connect the +ve poles of the batteries together and the -ve sides of the batteries together and add the resistor over the poles of bat 1. Now you have the batteries wired in parallell.
|----[ 10 kohm ]------|
|----[ - bat 1 + ] -----|
|----[ - bat 2 + ] -----|
You’ll have U = 10 V over the resistance, again giving you a current of I = U/R = 10/10000 = 1 mA. This current is split between the batteries, drawing 0.5 mA from each battery. At 0.5 mA, the batteries will last for (10/1000) / (0.5/1000) = 20 hours. The power generated will again be P = 10*1/1000 W = 10 mW.
Clear as mud?
Cheers,
Fred