PD across the terminals of two cells

[moenste]

Homework Statement

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?

 

[cnh1995]

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

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?

 

[moenste]

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?

Hm, PD = EMF + r?

 

[cnh1995]

Hm, PD = EMF - I*r

 

[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?

 

[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.

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

 

[moenste]

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

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...

 

[cnh1995]

@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?

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.

 

[cnh1995]

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...

Yes. That's the reason. Good job!

 

[cnh1995]

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

This is asking you the power supplied or absorbed by each cell.

(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.

What should be the condition for zero current through cell B?

 

[moenste]

This is asking you the power supplied or absorbed by each cell.

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

What should be the condition for zero current through cell B?

This one I'm struggling.

I tried to find the I_A: -3 + 5 (I_A - I_B) = 0 so I_A = 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 I_A = 0 so I_A = 0.83 A. So 10 - 0.83 R - 5 * 0.83 - 2 * 0.83 = 0 so R = 5.048 Ω.

What am I missing?

 

[cnh1995]

What am I missing?

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.

 

[moenste]

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.

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.

 

[gneill]

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.

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?

 

[moenste]

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?

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.

?

 

[gneill]

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

Okay, good.

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

?

Can you spell out what the above calculation is? What are each of the terms?

 

[moenste]

Can you spell out what the above calculation is? What are each of the terms?

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.

 

[gneill]

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.

 

[moenste]

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.

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?

 

[gneill]

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

 

[moenste]

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

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.

 

[gneill]

We can't find the PD between G and F. We need to know the resistance to find it.

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?

 

[moenste]

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:

View attachment 107153
What must ##V_x## be to prevent current from flowing in the loop?

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

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?

 

[gneill]

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.

 

[moenste]

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

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 VxV_x be to prevent current from flowing in the loop?

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

And how do we approach this one:

(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?

?

 

[gneill]

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

And how do we approach this one:

?

Right. So that's why I suggested starting with the required condition. You know that you want the potential drop across the GF portion of the potentiometer to balance the potential of battery B. And since you know the current passing through the potentiometer you can determine the resistance of the GF section.

 

[moenste]

Right. So that's why I suggested starting with the required condition. You know that you want the potential drop across the GF portion of the potentiometer to balance the potential of battery B. And since you know the current passing through the potentiometer you can determine the resistance of the GF section.

So this should be correct:
-3 - 3 * 0 + 0.83 R₁ = 0
R₁ = 3.6 Ω.

For the (f) part we need to find the position of F or in other words the resistances of the GF and FE parts right? But what is the correlation between the discharging and charging at the max. possible rates and the resistance?

 

[gneill]

For the (f) part we need to find the position of F or in other words the resistances of the GF and FE parts right? But what is the correlation between the discharging and charging at the max. possible rates and the resistance?

That you'll have to determine. What do you think is implied by maximum charging and maximum discharging? How does a battery charge or discharge?

 

[moenste]

That you'll have to determine. What do you think is implied by maximum charging and maximum discharging? How does a battery charge or discharge?

We had that topic in a different thread. I hope I got it right: a battery charges if it's current is going like - V is larger + ... + V is lower -. So the second battery is charging from the first one since it is pushing current to the second battery. And they are discharging if - V + ... - V +. In that case they work together and push current together.

 

[gneill]

We had that topic in a different thread. I hope I got it right: a battery charges if it's current is going like - V is larger + ... + V is lower -. So the second battery is charging from the first one since it is pushing current to the second battery. And they are discharging if - V + ... - V +. In that case they work together and push current together.

Okay, so discharging means current is flowing out of the + terminals of the battery, charging means current is flowing into the + terminal of the battery. Now you have to ponder how to maximize each of those with the given circuit. Remember, the only thing you can adjust is the position of the tap F on the potentiometer.

 

[moenste]

Okay, so discharging means current is flowing out of the + terminals of the battery, charging means current is flowing into the + terminal of the battery. Now you have to ponder how to maximize each of those with the given circuit. Remember, the only thing you can adjust is the position of the tap F on the potentiometer.

I think it all goes to the fact we need to put 10 Ω into one side. Like F should be placed either at G or at E.

If we put F at E:
10 - 10 I₁ + 10 I₃ - 2 I₁ = 0
-3 - 3 I₃ + 10 I₁ - 10 I₃ = 0
So I₃, flowing through B is 1.15 A.

If we put F at G:
10 - 12 I₁ = 0
I₁ = 0.83 A

-3 - 3 I₃ = 0
So I₃ = 1 A.

And so then we need to put F at E to make cell B charge at the maximum possible rate and put F at G to discharge B at maximum possible rate.

 

[gneill]

I agree with your reasoning as to where to place the potentiometer tap for maximizing charging or discharging.

 

[moenste]

I agree with your reasoning as to where to place the potentiometer tap for maximizing charging or discharging.

So it's correct? I was unsure about putting at G. And the calculations (at least the logic) are correct?

And in general: higher current flow is charging (better for charge), lower current flow is discharging (better for discharge)?

 

[gneill]

So it's correct? I was unsure about putting at G. And the calculations (at least the logic) are correct?

It looks okay. I calculate a slightly lower value for the charging current (1.14 A), but that could simply be a matter of rounding (I round down from 1.142 A).

And in general: higher current flow is charging (better for charge), lower current flow is discharging (better for discharge)?

Well, most people like their batteries to charge quickly but last a long time, so in that sense it's better. In real life one has to be careful about charging or discharging too quickly to prevent damaging the cell.

 

[gneill]

If you write your loop equations making the portion of the potentiometer shared by both loops a variable Rx (where Rx would range from 0 to 10 Ohms), then you can find a single equation for the B cell current that depends on the value of Rx, or equivalently, the position of the slider. With that you can plot a curve of the current versus position. Here's one that I just did, prettied up a bit to describe what's happening: