Current and resistance in the power loss of a transformer

[songoku]

Homework Statement

Diagram below shows a transformer having primary coil with resistance of 470 Ω and secondary coil with resistance 2950 Ω. The r.m.s potential difference across primary coil was maintained at 6 V. The efficiency of the transformer varies as R varies and when the transformer operates at maximum efficiency of 0.42, R is 13.5 kΩ. If the ratio of input and output voltage is 1 : 2, find:
a) the current in R
b) power loss in primary coil

Relevant Equations

$P=I^2 . R$

P = V.I

$P=\frac{V^2}{R}$

a) Output voltage = 2 x 6 = 12 V

Current in R = 12 V / 13.5 kΩ = 8.9 x 10^-4 A

That is the correct answer based on the solution but I don't understand why when calculating the current in R, the resistance of secondary coil is not considered. I thought it should be like this:

Current in R = 12 / (13.5 x 10³ + 2950) = 7.3 x 10^-4 A

Thanks

 

[haruspex]

Homework Statement:: Diagram below shows a transformer having primary coil with resistance of 470 Ω and secondary coil with resistance 2950 Ω. The r.m.s potential difference across primary coil was maintained at 6 V. The efficiency of the transformer varies as R varies and when the transformer operates at maximum efficiency of 0.42, R is 13.5 kΩ. If the ratio of input and output voltage is 1 : 2, find:
a) the current in R
b) power loss in primary coil
Relevant Equations:: $P=I^2 . R$

P = V.I

$P=\frac{V^2}{R}$

View attachment 287695

a) Output voltage = 2 x 6 = 12 V

Current in R = 12 V / 13.5 kΩ = 8.9 x 10^-4 A

That is the correct answer based on the solution but I don't understand why when calculating the current in R, the resistance of secondary coil is not considered. I thought it should be like this:

Current in R = 12 / (13.5 x 10³ + 2950) = 7.3 x 10^-4 A

Thanks

Are those two resistances in series as far as the voltmeter is concerned?
What is the potential across the external load?

 

[Delta2]

I am also tending towards the view that we should take the resistance of the secondary into account, however i am not so sure. The reason is (I think) from an energy/power point of view that $$V_sI_s=I_s^2R_s+I_s^2R\Rightarrow V_s=I_sR_s+I_sR$$

 

[haruspex]

I am also tending towards the view that we should take the resistance of the secondary into account, however i am not so sure. The reason is (I think) from an energy/power point of view that $$V_sI_s=I_s^2R_s+I_s^2R\Rightarrow V_s=I_sR_s+I_sR$$

So what do you think the right hand voltmeter reads, and how would you define "output voltage"?

 

[Delta2]

So what do you think the right hand voltmeter reads, and how would you define "output voltage"?

The right hand voltmeter says $\frac{12}{R_S+R}R$ and the output voltage is 12

 

[Delta2]

Tbh, I feel something just isn't quite right in what I am saying but not sure what it is. I think I am not sure if the voltage meter reads 12V or what I say in post #5.

 

[songoku]

What is the potential across the external load?

12 V

Are those two resistances in series as far as the voltmeter is concerned?

I thought so, but I also think that potential difference across secondary coil is the same as p.d. across the external load so maybe they are connected in parallel?

Thanks

 

[haruspex]

potential difference across secondary coil is the same as p.d. across the external load so maybe they are connected in parallel?

From the perspective of the voltmeter, yes.

 

[haruspex]

Tbh, I feel something just isn't quite right in what I am saying but not sure what it is. I think I am not sure if the voltage meter reads 12V or what I say in post #5.

If it does not refer to what the right hand voltmeter shows then it is not clear to me how output voltage is to be defined.

 

[songoku]

From the perspective of the voltmeter, yes.

So it means that the p.d. across coil is also 12 V

If I calculate the current through the secondary coil, it would be = 12 / 2950 = 4 x 10^-3 A

Why the current in secondary coil is not the same as current in R?

Thanks

 

[Delta2]

If it does not refer to what the right hand voltmeter shows then it is not clear to me how output voltage is to be defined.

What I have in mind is that the secondary of the transformer is like an EMF source with $E=12V$ and internal resistance $2950\Omega$

 

[haruspex]

What I have in mind is that the secondary of the transformer is like an EMF source with $E=12V$ and internal resistance $2950\Omega$

So it would be the open circuit voltage. Seems fair.
But that just makes the text ambiguous, and the given solution takes the other view.

 

[songoku]

Does it mean that there can be several interpretations for this question?

Also when calculating the power loss in primary coil, using P = I² . R and $P=\frac{V^2}{R}$ gives different result (just like the current I calculated in post 10)

Thanks

 

[Delta2]

Also when calculating the power loss in primary coil, using P = I2 . R and P=V2R gives different result (just like the current I calculated in post 10)

How do you calculate the current through the primary?

 

[songoku]

How do you calculate the current through the primary?

$$\eta = \frac{V_{s} I_{s}}{V_{p} I_{p}}$$
$$0.42=\frac{12 \times 8.9 \times 10^{-4}}{6 I_{p}}$$
$$I_p=4.2 \times 10^{-3} \text{A}$$

Power loss in primary coil = $(4.2 \times 10^{-3})^{2} \times 470 = 8.3 \times 10^{-3}~\text{W}$

Using $P=\frac{V^2}{R}$ , power loss = $\frac{6^2}{470}=0.08 ~ \text{W}$

I am quite confused with this question. Nothing I calculate match each other, either the power loss or current through secondary coil and external load

Thanks

 

[Delta2]

I think your first method is correct, the second method doesn't take into account that the primary current depends on the current of the secondary , due to the mutual inductive coupling of the two coils.

So our only problem now is what exactly is the value of current on the secondary...

 

[songoku]

the second method doesn't take into account that the primary current depends on the current of the secondary , due to the mutual inductive coupling of the two coils.

But is it correct for me to say that the p.d. across the primary coil is 6 V? Because this is the value I used to calculate current in primary coil.

If yes, then why $P=\frac{V^2}{R}$ does not work? I calculate the power in primary coil using the voltage on primary coil and resistance of primary coil.

Thanks

 

[Delta2]

If yes, then why P=V2R does not work? I calculate the power in primary coil using the voltage on primary coil and resistance of primary coil.

You just can't use plain Ohm's law to calculate the current through the primary. Two reasons

1. Because that way you don't use the given efficiency 0.42 (ok this sounds funny I know)
2. From what I know the full ODE for the current through the primary is $$V(t)-I_pR_p-L_p\frac{dI_p}{dt}-M\frac{dI_s}{dt}=0$$ (M mutual inductance, $L_p$ self inductance of primary, V(t) source voltage) and it seems to me that Ohm's law is an oversimplification of that equation (it completely neglects the effect of Mutual and self inductance)

 

[Steve4Physics]

Hi, I’m chipping-in if I may...

a) The output voltage (between the secondary's terminals) is 12V - because we are told it is twice 6V.

The current through R is therefore $I = \frac V R = \frac {12}{13.5\times10^3} = 8.9 \times10^{-4} A$
__________
Side-note

The secondary coil and the load have the same current, so the voltage-drop (‘lost volts’)across the secondary coil’s internal resistance (r) is $Ir = 8.9 \times 10^{-4} \times 2950 = 2.6V.$

That means the secondary’s emf $= 12+2.6 = 14.6$V.
__________

b) There is no need to find the primary current (or even use the primary’s resistance).

The power delivered to R is $\frac {V^2}{R} = \frac {12^2}{13.5\times10^3} = 0.0107W$

The tranformer’s efficiency is 0.42. That means the input power to the transformer is $\frac {0.0107}{0.42} = 0.0254$W

From conservation of energy, the power loss in the primary coil is then:
(input power) – (power dissipated in R) – (power dissipated in secondary coil)

 

[songoku]

Side-note

The secondary coil and the load have the same current, so the voltage-drop (‘lost volts’)across the secondary coil’s internal resistance (r) is $Ir = 8.9 \times 10^{-4} \times 2950 = 2.6V.$

That means the secondary’s emf $= 12+2.6 = 14.6$V.
__________

b) There is no need to find the primary current (or even use the primary’s resistance).

The power delivered to R is $\frac {V^2}{R} = \frac {12^2}{13.5\times10^3} = 0.0107W$

The tranformer’s efficiency is 0.42. That means the input power to the transformer is $\frac {0.0107}{0.42} = 0.0254$W

From conservation of energy, the power loss in the primary coil is then:
(input power) – (power dissipated in R) – (power dissipated in secondary coil)

So my first method in post #15 is wrong since I use 12 as V_s , where I should use 14.6 V?

I tried to redo it:
$$\eta =\frac{V_s I_s}{V_p I_p}$$
$$0.42 =\frac{14.6 \times 8.9 \times 10^{-4}}{6 \times I_p}$$
$$I_p=5.2 \times 10^{-3}~\text{A}$$

Power loss in primary coil = $(5.2 \times 10^{-3})^2 \times 470=0.0127 ~W$

This result can be said the same as the one obtained from your method:
power loss in primary coil = input power - power dissipated in R - power dissipated in secondary coil
= 0.0254 - 0.0107 - (2.6² / 2950)
= 0.0124 W

But the given solution is 8.3 x 10^-3 W

Maybe I will just assume the solution is wrong

 

[Steve4Physics]

This result can be said the same as the one obtained from your method:
power loss in primary coil = input power - power dissipated in R - power dissipated in secondary coil
= 0.0254 - 0.0107 - (2.6² / 2950)
= 0.0124 W

But the given solution is 8.3 x 10^-3 W

Maybe I will just assume the solution is wrong

The official answer to part a) (8.9x10⁻⁴A) is calculated by taking the secondary’s terminal voltage as 12V. (This means the emf = 14.6V.)

But the official answer to part b) (8.3x10⁻³W) has been calculated assuming the emf is 12V (which would make the secondary’s terminal voltage less than 12V).

That's what's gone wrong.

So I would say the official answer to part b) is wrong. The incorrect answer (8.3x10⁻³W) can be calculated, by (incorrectly) taking the emf as 12V, as follows:

Current in secondary $I = \frac {V}{R+r} = \frac {12}{2950+13500} = 0.0007295A$

Power loss in secondary coil $= I^2r = 0.0007295^2 \times 2950 = 0.00157W$
Power delivered to R $= I^2R = 0.0007295^2 \times 13500 = 0.007184W$
Input power to primary $= \frac {0.007184}{0.42} = 0.0171W$

Power loss in primary coil $= 0.0171 – 0.00157 – 0.007184 = 0.0083W$

(And note, there is no need to use the primary’s resistance unless you are also required to find the primary current. Conservation of energy makes the calculation much simpler!)

Edit: Of course it could be the other way round: the official answer to part a) is wrong and the answer for part b) is correct.

 

[songoku]

Thank you very much for all the help Delta2, haruspex, Steve4Physics