# Homework Help: PD across the terminals of two cells

1. Oct 7, 2016

### moenste

1. The problem statement, all variables and given/known data
In the following circuit, cell A has an EMF of 10 V and an internal resistance of 2 Ω; cell B has an EMF of 3 V and an internal resistance of 3 Ω.

(a) Show that the currents through A and B are 65 / 71 amps and 14 / 71 amps respectively. What is the magnitude of the current through GF?

(b) Determine the power dissipated as heat in the resistor FE. If the circuit is switched on for 30 minutes, calculate the energy dissipated in FE in kilowatt-hours.

(c) What is the potential difference (i) across the terminals of cell A, (ii) across the terminals of cell B?

(d) Calculate the rate at which energy is being supplied (or absorbed) by cells A and B.

(e) If the contact F can be moved along the resistor GE, find the value of the resistance GF when no current is flowing through cell B.

(f) At what setting of F would cell B (i) be discharging at the maximum possible rate and (ii) be charging at the maximum possible rate?

Answers: (a) 0.718 A, (b) 4.19 W, 2.095 * 10-3 kWh, (c) 8.17 V, 3.59 V, (d) 9.15 W, 0.592 W, (e) 3.60 Ω.

2. The attempt at a solution
(a) 10 - 5 (I1 - I3) - 5 I1 - 2 I1 = 0 → I3 = 2.4 I1 - 2. Plug into: -3 -3 I3 + 5 I1 - 5 I3 = 0 → I1 = 0.9155 A (65 / 71), I3 = 0.197 A and I2 = 0.718 A.

(b) P = V I and V = R I = 5 * (65 / 71) = 4.6 V. Then P = 4.6 * (65 / 71) = 4.19 W.
W = V I t = 4.6 * (65 / 71) * (30 * 60) = 7543 J. 1 kWh = 3.6 * 106 J so 7543 / 3.6 * 106 = 2.095 * 10-3 kWh.

(c) This is where difficulties come. As I understand, I should calculate the voltage of the circuits. (i) V = I R = (65 / 71) * (5 + 2) = 6.4 V of the 5 Ω and 2 Ω resistors around A. (ii) We calculte the voltage of the 3 Ω resistor: V = (14 / 71) * 3 = 0.59 V plus 3 V is 3.59 V for the 5 Ω resistor since it is parallel.

I didn't continue since I am not sure on (c). What's wrong with it?

2. Oct 7, 2016

### cnh1995

You are asked to find the terminal voltages of the cells, given the current, emf and internal resistance of each cell. What is the formula for terminal voltage (p.d.) of the cell then?

3. Oct 7, 2016

### moenste

Hm, PD = EMF + r?

4. Oct 7, 2016

5. Oct 7, 2016

### moenste

@cnh1995
I get the correct answer for (c) (ii): 3 + 3 * (14 / 71) = 3.59 V.

But what's wrong with (c) (i): 10 + 2 * (65 / 71) = 11.8 V?

6. Oct 7, 2016

### cnh1995

Sorry I edited my previous post. Forgot to replace + sign with -.

7. Oct 7, 2016

### moenste

Hm, in that case I get a correct answer for (i) and a wrong one for (ii) .

I did a drawing for (a) and (b) and in my drawing a have current going from A to the left and then entering B from the left and then current enters A from the right. So in terms of B we have + 3 V - ... + 3 Ω - and so we have PD = EMF + I r. And in A we have + 10 V - ... - 2 Ω + and so we PD = EMF - I r.

I think this is the correct reason...

8. Oct 7, 2016

### cnh1995

You need to consider the directions of the currents. What is the direction of the current through B? Terminal voltage of B is greater than the emf.

9. Oct 7, 2016

### cnh1995

Yes. That's the reason. Good job!

10. Oct 7, 2016

### cnh1995

This is asking you the power supplied or absorbed by each cell.
What should be the condition for zero current through cell B?

11. Oct 7, 2016

### moenste

This one I solved: P = V I → PA = 10 * (65 / 71) = 9.15 W and PB = 3 * (14 / 76) = 0.59 W.

This one I'm struggling.

I tried to find the IA: -3 + 5 (IA - IB) = 0 so IA = 0.6 A. Then plug it in 10 - 0.6 R - 5 * 0.6 - 2 * 0.6 = 0 so R = 9.6 Ω.

I also tried just to do 10 - (65 / 71) R - 5 (65 / 61) - 2 (65 / 71) = 0 so R = 3.92 Ω.

Also tried 10 - 12 IA = 0 so IA = 0.83 A. So 10 - 0.83 R - 5 * 0.83 - 2 * 0.83 = 0 so R = 5.048 Ω.

What am I missing?

12. Oct 7, 2016

### cnh1995

The condition for zero current through cell B.
What should be the voltage across GF? Or in simple words, what should be the voltage across the internal resistance of cell B? Consider only the loop containing G-F resistance and cell B. Draw this loop and see.

Last edited: Oct 7, 2016
13. Oct 8, 2016

### moenste

If no current is flowing through cell B then all current is flowing through the GF resistor. And then we have R = V / I = 3.59 V / (65 / 71) = 3.92 Ω.

14. Oct 8, 2016

### Staff: Mentor

By moving the slider on the variable resistor (also known as a potentiometer), you're changing the circuit. So the old values of current and potential differences no longer hold. You need to re-evaluate the circuit for the new conditions.

Since you're looking for conditions where cell B has zero current flowing, start by imagining that the lower cell is not connected at F (open that connection temporarily). This will let you evaluate what will be going on in the upper loop under those conditions. What then is the current in the upper loop?

15. Oct 8, 2016

### moenste

I3 = 0 so 10 - 5 (I1 - I3) - 5 I1 - 2 I1 = 0 so I1 = 0.833 A.

And then 10 - 0.83 R - 5 * 0.83 - 2 * 0.83 = 0 so R = 5 Ω.

?

16. Oct 8, 2016

### Staff: Mentor

Okay, good.
Can you spell out what the above calculation is? What are each of the terms?

17. Oct 8, 2016

### moenste

This is the same circuit but with the unknown GF resistor of 5 Ω which we need to find (the AGEA area).

But in that case we'll get the same answer...

We can't go through GBFG though. I mean GBF is zero.

18. Oct 8, 2016

### Staff: Mentor

The resistor GFE has a total resistance (from G to E) of 10 Ω. When the slider F moves, it divides the resistor into two parts, the sum of which remains 10 Ω. So for example, if F is located exactly in the middle then the two parts will be 5 Ω each. But if it's moved away from the middle they might be 2 Ω and 8 Ω, or 7.3 Ω and 2.7 Ω, or any combination that sums to 10 Ω. What you need to do is determine what combination will yield the desired condition that when cell B is connected at their junction, no current flows in the lower loop.

Hint: you should recognize that GFE along with the 10 V battery forms a potential divider.

19. Oct 8, 2016

### moenste

10 = 0.83 R1 + 0.83 R2 + 0.83 * 2
R2 = 10.048 - R1

That's as far as I can go.

Update:
-3 - 3 * 0 + 0.83 R1 = 0
R1 = 3.6 Ω

It gets the corect answer. But how can we consider it, since the whole GBF line is zero?

20. Oct 8, 2016

### Staff: Mentor

Start with the required condition that prevents current from flowing in the bottom loop. What must the potential difference be between G and F?

21. Oct 8, 2016

### moenste

We can't find the PD between G and F. We need to know the resistance to find it. We can only say that V = I R = 0.83 * 2 = 1.66 V is the PD of the internal resistor and the GE line has 10 - 1.66 = 8.34 V.

And if we put 8.34 = 0.83 R1 + 0.83 R2 we'll get the same result as in post # 19.

22. Oct 8, 2016

### Staff: Mentor

That's not true. You can specify the potential difference first, then calculate the required resistance. What potential difference will prevent current from flowing in the bottom loop?

Consider the following example where you have a known battery of 3 V with some internal resistance, and some other (unknown) potential difference source:

What must $V_x$ be to prevent current from flowing in the loop?

23. Oct 8, 2016

### moenste

I think you are suggesting this (maybe you missed it, I updated the post later):

24. Oct 8, 2016

### Staff: Mentor

It works because if the bottom loop has no current flowing then the upper loop is effectively isolated and its current is fixed solely by the resistance in its path: the internal resistance of the 10 V battery and the 10 Ohms of the potentiometer. Then you simply need to find the potential difference across R1 that makes your equation true. R1 has the upper loop's current running through it all alone, since the lower loop is contributing no current.

The situation in the example in post #22 shows what potential difference is required to prevent current flow.

25. Oct 8, 2016

### moenste

VX - Ir - 3 = 0
VX = 3 V since I = 0.

And how do we approach this one:
?