The actual current flowing is 0.137A. The sum of internal and external resistance is
R = 30V/.137A = 219 ohms
This equals 200 + r1 + r2 = r3
= 200 + b(E1+E2+E3) = 200 + 30b
Therefore 19 ohms = 30b
Solve for b, in ohms/volt.
Batteries are not perfect. Instead they have an "internal resistance, a catch all phrase which describes the internal chemical workings of the battery which prevents it from providing an unchanging voltage regardless ofnits load. A simple model of internal resistance attaches a small resistor to each battery in a circuit. Consider a circuit with three batteries connected in series to a 200 ohms resistor.
Lets assume the internal resistance gets worse as the battery provides a greater EMF. To that end , lets see the internal resistance r(i) of the battery with EMF E(i) equal to be bE(i) where b is a constant assumed the same for all the three batteries. If the battteries have EMF's 5v, 10v, 15v and there is 0.137 A of current flowing from the battery ,find the value of b.
(E1=5v;E2=10v;E3=15v;
r1=bE1;r2=bE2;r3=bE3
I=.137 A)
1 answer
1 answer