Internal resistance of the battery and energy transformed

In summary: Watt is a person's name, not a unit), which is the rate of energy dissipation. The Joule is a unit of energy and is defined as a unit group (N m) where the Newton is a unit of force and the meter is a unit of distance. So a Watt is a Newton-meter per second, which is the same as a Joule per second. So your W = V I t equation gives you the energy delivered (or consumed) in t seconds.
  • #1
moenste
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Homework Statement


The battery in the circuit below has EMF 5.4 V and drives a current of 0.30 A through a lamp. The voltmeter reading is 4.8 V.

8da6ce6eef61.jpg


Explain why the voltmeter reading is less than the EMF of the cell.

Calculate values for (a) the internal resistance of the battery, and (b) the energy transformed per second in the lamp.

State two assumptions you made in order to complete these calculations.

Answers: (a) 2.0 Ω, (b) 1.44 W

2. The attempt at a solution
The voltmeter reading is less than the EMF of the cell because the EMF is a non-ideal power source that has internal resistance and therefore the output 4.8 V is lower than the input 5.4 V.

(a) E = I r + V → 5.4 V = 0.3 A * r + 4.8 V → r = 2 Ω.

(b) W = V I t = 4.8 V * 0.3 A * 1 s = 1.44 W.

I would assume that the ammeter has no resistance and therefore does not disturb the circuit. I would also assume that the voltmeter has infinite resistance and also does not disturb the circuit.

I am mostly unsure on the theory part. Is it correct? What's wrong?
 
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  • #2
moenste said:

Homework Statement


The battery in the circuit below has EMF 5.4 V and drives a current of 0.30 A through a lamp. The voltmeter reading is 4.8 V.

8da6ce6eef61.jpg


Explain why the voltmeter reading is less than the EMF of the cell.

Calculate values for (a) the internal resistance of the battery, and (b) the energy transformed per second in the lamp.

State two assumptions you made in order to complete these calculations.

Answers: (a) 2.0 Ω, (b) 1.44 W

2. The attempt at a solution
The voltmeter reading is less than the EMF of the cell because the EMF is a non-ideal power source that has internal resistance and therefore the output 4.8 V is lower than the input 5.4 V.

(a) E = I r + V → 5.4 V = 0.3 A * r + 4.8 V → r = 2 Ω.

(b) W = V I t = 4.8 V * 0.3 A * 1 s = 1.44 W.

I would assume that the ammeter has no resistance and therefore does not disturb the circuit. I would also assume that the voltmeter has infinite resistance and also does not disturb the circuit.

I am mostly unsure on the theory part. Is it correct? What's wrong?
It all looks correct to me.
 
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  • #3
moenste said:
Calculate values for the energy transformed per second in the lamp.
moenste said:
W = V I t = 4.8 V * 0.3 A * 1 s = 1.44 W.
Voltage x current x time = energy.
The question asks you to find energy, so you did use the right equation. But you have to change the units of your answer. Also, the given data has a precision of 2 significant figures, which should be reflected in your answer.
 
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  • #4
David Lewis said:
Voltage x current x time = energy.
The question asks you to find energy, so you did use the right equation. But you have to change the units of your answer. Also, the given data has a precision of 2 significant figures, which should be reflected in your answer.
"W = the energy dissipated in some time t." -- this is what I have in my book regarding "energy and power in DC circuits".

My derived answer is 1.44 W, my book answer is 1.44 W. I'm not sure whether I got you correctly... Maybe you mean I should have my (a) answer 2.00 Ω instead of 2 Ω? Otherwise I've no idea, since I have two digits (.44) and have watts (1 W = 1 J s-1).
 
  • #5
moenste said:
(b) W = V I t = 4.8 V * 0.3 A * 1 s = 1.44 W.

moenste said:
My derived answer is 1.44 W, my book answer is 1.44 W. I'm not sure whether I got you correctly... Maybe you mean I should have my (a) answer 2.00 Ω instead of 2 Ω? Otherwise I've no idea, since I have two digits (.44) and have watts (1 W = 1 J s-1).
It's not a matter of significant figures but rather of keeping track of units. Volts x Amps = Watts, and a Watt is a Joule per second. So Watts gives you energy per second, which is what was asked for.

If you multiply Watts by a time value (like your 1 s) then you end up with the energy delivered (or consumed) over that particular time span in joules, not joules per second. So what is being pointed out is that you should not have multiplied by the 1 s value as it makes the math incorrect unit-wise.
 
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  • #6
gneill said:
It's not a matter of significant figures but rather of keeping track of units. Volts x Amps = Watts, and a Watt is a Joule per second. So Watts gives you energy per second, which is what was asked for.

If you multiply Watts by a time value (like your 1 s) then you end up with the energy delivered (or consumed) over that particular time span in joules, not joules per second. So what is being pointed out is that you should not have multiplied by the 1 s value as it makes the math incorrect unit-wise.
I understand that. Why is the formula then W = V I t?

W = Q V
Q = I t
so W = V I t

Update: W is the energy dissipated in some time t and P is the rate of dissipation of energy (or power). P = V I = x Watt.

I think I got it. Thank you!
 
  • #7
moenste said:
I understand that. Why is the formula then W = V I t?

W = Q V
Q = I t
so W = V I t

Update: W is the energy dissipated in some time t and P is the rate of dissipation of energy (or power). P = V I = x Watt.

I think I got it. Thank you!
Okay. Just to be sure, the Watt is the name given to the unit group J/s (that is, joules per second). It is the rate at which energy is changing, also known as power. In fact the Joule is also a compounded unit! ##Joule = kg~m^2~s^{-2}##. Energy and power are such useful concepts that they were given their own unit names.
 
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  • #8
gneill said:
Okay. Just to be sure, the Watt is the name given to the unit group J/s (that is, joules per second). It is the rate at which energy is changing, also known as power. In fact the Joule is also a compounded unit! ##Joule = kg~m^2~s^{-2}##. Energy and power are such useful concepts that they were given their own unit names.
In our case we need to find it per 1 sec, so we use P = V I = x W (per second).

And W = V I t or W = P t is the rate of work per time X. Right?
 
  • #9
moenste said:
In our case we need to find it per 1 sec, so we use P = V I = x W (per second).

And W = V I t or W = P t is the rate of work per time X. Right?
No. Watts is the answer. Energy per second, that's what you were asked to find. From the problem statement:
moenste said:
(b) the energy transformed per second in the lamp.
Note that it doesn't say "in one second", it says "per second". The nuances of precise technical language can be a bit tricky at times :smile:
 
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  • #10
gneill said:
No. Watts is the answer. Energy per second, that's what you were asked to find. From the problem statement:

Note that it doesn't say "in one second", it says "per second". The nuances of precise technical language can be a bit tricky at times :smile:
Got it now :).
 
  • #11
moenste said:
Calculate the energy transformed per second in the lamp.
I may be wrong, but my answer is 1.4 J. The problem asks for energy, so that is what I give them.
Also, my answer is rounded to 2 significant figures.
 
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  • #12
David Lewis said:
I may be wrong, but my answer is 1.4 J. The problem asks for energy, so that is what I give them.
Also, my answer is rounded to 2 significant figures.
Hm, but isn't it asking for energy per second? And it is 1 W = 1 J s-1?

And P = V * I = 4.8 V * 0.3 A = 1.44 W, my phone (which is not a scientific calculator) also gives 1.44 not 1.4...

So I think 1.44 W or 1.44 J s-1 should be correct.
 
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  • #13
moenste said:
So I think 1.44 W or 1.44 J s-1 should be correct.
I agree..
 
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  • #14
gneill said:
Note that it doesn't say "in one second", it says "per second".
I don’t understand the distinction.
 
  • #15
David Lewis said:
I don’t understand the distinction.
Perhaps it's a matter of interpretation of the nuances of natural language description of mathematical concepts. Or perhaps just understanding what the author of the question intends. Or perhaps it's the experience of having dealt with this type of question many times :smile::smile:

When a scientist asks for the energy produced in a specific amount of time he's asking for a particular number in energy units (Joules in this case) for that particular interval of time. When he asks for the energy produced per second, it implies an ongoing process and power (Watts) is implied. I realize it's a nebulous, perhaps dubious technicality, but it's the sort of thing that recurs over and over again in problem sets.
 
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1. What is internal resistance of a battery?

The internal resistance of a battery is the resistance that exists within the battery itself, which affects the flow of electric current. It is caused by the resistance of the materials used in the battery's construction and the chemical reactions that occur within the battery.

2. How does internal resistance affect the performance of a battery?

The internal resistance of a battery can affect its performance in several ways. It can cause voltage drops, decrease the battery's overall capacity, and increase the battery's self-discharge rate. It can also cause the battery to heat up, which can lead to a decrease in its lifespan.

3. How is internal resistance measured?

Internal resistance is typically measured by using a multimeter to measure the voltage drop across the battery while a known current is flowing through it. The internal resistance can then be calculated using Ohm's Law (R=V/I), where R is the internal resistance, V is the voltage drop, and I is the current.

4. Can internal resistance be reduced?

While internal resistance is a natural characteristic of batteries, it can be reduced by using materials with lower resistance in the battery's construction and optimizing the chemical reactions within the battery. However, reducing internal resistance can also lead to a decrease in the battery's overall capacity.

5. How does internal resistance relate to energy transformation in a battery?

The internal resistance of a battery affects the energy transformation process by converting some of the battery's energy into heat, rather than usable electrical energy. This can lead to a decrease in the battery's overall efficiency and the amount of energy that can be transformed from chemical energy to electrical energy.

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