# Internal resistance of the battery and energy transformed

## 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.

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?

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
cnh1995
Homework Helper
Gold Member

## 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.

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.

moenste
Calculate values for the energy transformed per second in the lamp.
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.

moenste
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).

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

moenste
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!

gneill
Mentor
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.

moenste
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?

gneill
Mentor
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:
(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

moenste
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
Got it now :).

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.

moenste
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.

gneill
gneill
Mentor
So I think 1.44 W or 1.44 J s-1 should be correct.
I agree..

moenste
Note that it doesn't say "in one second", it says "per second".
I don’t understand the distinction.

gneill
Mentor
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

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.

moenste and David Lewis