A battery is made of 3cells in series. Each cell hasε= 2.20V, r = 0.0100Ω. The battery is connected to a 3.27Ωload. Calculate the potential difference across the load.

Answer 1

I = E/(r+Rl) = 3*2.2/(0.01+3.27) = 2.012A.
V = I * Rl = 2.012 * 3.27 =

Answer 2

Same type of problem, different names? Huummm

Answer 3

total ε is ... 3 * 2.20 v

total r is ... 3.27Ω + (3 * 0.0100Ω)

find circuit current

calculate load potential

Answer 4

Correction:
I = 6.6/(0.03+3.27) = 2A.
V = I * Rl = 2 * 3.27 =