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)

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.

To find the value of b in this scenario, we can use Ohm's Law and the concept of series resistance.

In a series circuit like this one, the total resistance is the sum of the individual resistances. So, the total resistance in this circuit is:

R_total = r1 + r2 + r3 + R_load

Where R_load is the resistance of the load, which is given as 200 ohms.

Now, we can also use Ohm's Law to express the current flowing through the circuit:

I = (E1 + E2 + E3) / (r1 + r2 + r3 + R_load)

Substituting the values given in the question, we have:

0.137 A = (5 V + 10 V + 15 V) / (r1 + r2 + r3 + 200 ohms)

Simplifying the equation:

0.137 A = 30 V / (bE1 + bE2 + bE3 + 200 ohms)

0.137 A = 30 V / (b (E1 + E2 + E3) + 200 ohms)

0.137 A = 30 V / (b (5 V + 10 V + 15 V) + 200 ohms)

0.137 A = 30 V / (30 b V + 200 ohms)

0.137 A = 1 V / (b + 6.67 ohms)

Now, to solve for b, we need to isolate it. We can achieve this by rearranging the equation:

1 V = (0.137 A) * (b + 6.67 ohms)

1 V / 0.137 A = b + 6.67 ohms

7.3 ohms = b + 6.67 ohms

b = 7.3 ohms - 6.67 ohms

b = 0.63 ohms

Therefore, the value of b is 0.63 ohms.