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.
A)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)
B) To decrease the internal resistance , we can connect the batteries in parallel;however our advantage may be offset by a decrease in net voltage.To test this hypothesis, find the current flowing through the 200 ohms resistor, which is connected i series with the three parallel batteries.Assume that b has the same value as in part (a). Comment on the whether its worth connecting these batteries in parallel if the gosl is to send as much current as possible through the 200 ohms resistor.
Note: I just wanna know how to solve the B part....
I already did part A for you.
In part B, you NEVER connect batteries of unequal voltage in parallel, otherwise the higher voltage battery will discharge through the lower voltage batteries. In your case, you might consider using the 5 and 10 V batteries in series, and putting them in parallel with the 15 V battery. this will not result in any more current through the 200 ohm resistor, since the overall voltage applied to the resistor cannot exceed 15V. Even with no internal resistance, the current would be 15/200 = 0.075 A, which is less than before when they were in series.
To solve part B of the problem, we can follow these steps:
1. We know that the internal resistance of each battery is given by r(i) = bE(i), where b is a constant and E(i) is the EMF of the battery.
2. In part B, we are connecting the batteries in parallel and then connecting them in series with the 200 ohms resistor.
3. When batteries are connected in parallel, the total internal resistance (r(t)) is given by the reciprocal of the sum of the reciprocals of the individual internal resistances:
1/r(t) = (1/r1) + (1/r2) + (1/r3)
4. Since r(i) = bE(i), we have:
1/r(t) = (1/bE1) + (1/bE2) + (1/bE3)
Simplifying this equation, we get:
1/r(t) = (E2E3 + E1E3 + E1E2) / (bE1E2E3)
5. Now, we can find the current flowing through the 200 ohms resistor by using Ohm's Law:
I = V / R
In this case, V is the net voltage from the batteries connected in parallel, and R is the resistance of the 200 ohms resistor.
6. The net voltage (V) is given by the sum of the EMFs of the batteries (E1 + E2 + E3) minus the voltage dropped across the internal resistances (I * r(t)):
V = (E1 + E2 + E3) - (I * r(t))
7. Substitute the value of 1/r(t) from step 4 into the equation V = (E1 + E2 + E3) - (I * r(t)).
8. Solve for I (the current flowing through the 200 ohms resistor). This will give you the answer to part B of the problem.
9. Finally, compare the calculated value of I with the current flowing through the resistor when the batteries are connected in series from part A. If the calculated value of I is higher, then it's worth connecting the batteries in parallel to maximize the current flowing through the 200 ohms resistor. Otherwise, it may not be worth it.