How Do Transformers Affect Voltage, Current, and Light Bulb Brightness?

[IanT]

I am confused over something about voltage and current. A transformer increases/decreases the voltage in a circuit and does the opposite to the current. The current times the voltage has the same value before and after. This however, seems to go against the well known equation V = IR (the voltage equals the current times the resistance). If a step-up transformer increases the voltage and decreases the current, doesn't the resistance have to increase A LOT in order for V = IR to be valid?

Another thing is how the voltage affects the brightness of a light bulb. My textbook says that a light bulb connected to a circuit after a step-up transformer is brighter than one connected before it. But although the voltage is higher, the current is lower, meaning that the joules per second before and after going through a step-up transformer is the same. So why is the light bulb brighter?

Thanks in advance.

 

[Andrew Mason]

I am confused over something about voltage and current. A transformer increases/decreases the voltage in a circuit and does the opposite to the current.

A transformer transforms voltage not current. It is true that the current in the primary is related to the current in the secondary by: $V_pI_p = V_sI_s$. But this simply means that the secondary current determines the primary current. $I_s = V_s/R => I_p = V_sI_s/V_p = V_s^2/RV_p$ where R is the resistance of the secondary (assuming all loads are resistive). In other words, for a given primary voltage, the current in both the primary and secondary is determined by the load in the secondary.

Another thing is how the voltage affects the brightness of a light bulb. My textbook says that a light bulb connected to a circuit after a step-up transformer is brighter than one connected before it. But although the voltage is higher, the current is lower, meaning that the joules per second before and after going through a step-up transformer is the same. So why is the light bulb brighter?

Again, the transformer transforms voltage, not current. If the voltage applied to the light bulb increases, the current will increase proportionately according to Ohm's law: I = V/R.

AM

 

[IanT]

I'm sorry but I don't understand.

Let's say that the voltage is 1V and the current is 2A in the primary circuit and the voltage is 2 V and the current is 1A in the secondary circuit. This is consistent with Vp*Vs = Vs*Is.

Since V = IR, in the primary circuit, 1V = 2A*R and in the secondary cicuit, 2V = 1A*R. In order for this to be true, the resistance, R, has to be half an ohm in the first circuit and 2 ohms in the second. Does this happen? Does the resistance increase?"Again, the transformer transforms voltage, not current. If the voltage applied to the light bulb increases, the current will increase proportionately according to Ohm's law: I = V/R. "

But how does it increase proportionally if Vp*Vs = Vs*Is? If the voltage applied increases, the current has to decrease according to that. One light bulb is connected to the primary circuit and the other to the secondary circuit.

 

[yster]

You should do a little bit of reading on open circuit and closed circuit current of a transformer. Andrew explained this very well.

The transformer transforms Voltage, not current. The current is dependent on the load in the secondary side.

A good way to approach it is. 1st determine what the secondary voltage will be then determine what the secondary current based on the load and this voltage then determine the primary current and all will make sense.

Don't try to determine both current and voltage at the same time: You have the right equations but you're apllying them wrong

Look again at Andrews post all the equations are there.

 

[Jobrag]

If you put the bulb across the primary connections you would get Ip=Vp/R
Your transformer is 2:1 step up so if you now connect the bulb to the secondary terminals you get Is=2*Vp/R so Is=2*Ip.
In the first case the power taken from the system is Vp*Ip, in the second case the power taken from the system is Vs*Is (Vs=2*Vp Is=2*Ip) so the power in the second case is 4*Vp*Ip.
In other words when you connect your bulb to the secondary side Ip is eour times what it would be if connected to the primary. Lot hotter lot brighter.

 

[Andrew Mason]

I'm sorry but I don't understand.

Let's say that the voltage is 1V and the current is 2A in the primary circuit and the voltage is 2 V and the current is 1A in the secondary circuit. This is consistent with Vp*Vs = Vs*Is.

Since V = IR, in the primary circuit, 1V = 2A*R and in the secondary cicuit, 2V = 1A*R. In order for this to be true, the resistance, R, has to be half an ohm in the first circuit and 2 ohms in the second. Does this happen? Does the resistance increase?

This is not correct for the primary circuit. The current in the primary is limited by inductive reactance. If the primary coil was a perfect conductor, there would be no resistance at all. The load in the secondary determines the current in the secondary and that current affects the inductive reactance in the primary circuit. The greater the current in the secondary, the lower the inductive reactance in the primary.

AM

 

[IanT]

Thanks for the replies. I don't have a problem with solving any problems from my textbook, I can usually do it with one equation or the other. I was just confused over an apparent contradiction between two equations. I think I'm understanding now.

Can somebody explain more about how inductive reactance works? I don't believe I was taught that.

Again, thanks for going out of your way to help an internet stranger.

 

[Andrew Mason]

Thanks for the replies. I don't have a problem with solving any problems from my textbook, I can usually do it with one equation or the other. I was just confused over an apparent contradiction between two equations. I think I'm understanding now.

Can somebody explain more about how inductive reactance works? I don't believe I was taught that.

Again, thanks for going out of your way to help an internet stranger.

It is great to see that you are interested and have these questions. Why not do some reading on your own ask your teacher first?

AM

 

[IanT]

I don't have a teacher and the wikipedia article on reactance is quite inaccessible. Maybe you could link me somewhere with a better explanation for it?

I still don't get the answer for the second question from the OP. The voltage is transformed, but the power input = power output. Why is a light bulb connected to the primary dimmer than one connected to the secondary?

 

[Andrew Mason]

I don't have a teacher and the wikipedia article on reactance is quite inaccessible. Maybe you could link me somewhere with a better explanation for it?

Try http://nsdl.org/resource/2200/20100509220739292T. Chapters 14 and 15 deal with inductance.

I still don't get the answer for the second question from the OP. The voltage is transformed, but the power input = power output. Why is a light bulb connected to the primary dimmer than one connected to the secondary?

It is just Ohm's law. The voltage across the primary is lower than the voltage across the secondary (because it is a step-up transformer). So the current (I = V/R) is higher.

AM

 

[IanT]

I = V/R implies that the current changes proportionally with the voltage. You would expect the current to be lower if the voltage is lower. Can I have an example of a set up where the voltage, current and resistance (if any) is known for both the primary and secondary? I don't see how this can be possible without contradicting either ohm's law (I = V/R) or the relationship between the current and voltage of the primary and secondary. (Vp*Ip = Vs*Is) I was under the impression than there was some sort of exception for the primary because of "inductive reactance".

 

[Jobrag]

I still don't get the answer for the second question from the OP. The voltage is transformed, but the power input = power output. Why is a light bulb connected to the primary dimmer than one connected to the secondary?

Because when you are connected to the secondary you are drawing four times the amount of power as when you are connected across the primary.

Leave the transformer out of it for the moment.
You connect a lightbulb resistance R across a 1 V supply I=V/R so I= 1/R Power = I*V = 1*1/R = 1/R

Now connect the bulb across a 2V supply I=2/R Power = 2*2/R=4/R
You are happy that a lightbulb consuming four times the power will shine more brightly?
For all the pedants; yes I know it should be a lamp not a lightbulb.

 

[Andrew Mason]

I = V/R implies that the current changes proportionally with the voltage. You would expect the current to be lower if the voltage is lower. Can I have an example of a set up where the voltage, current and resistance (if any) is known for both the primary and secondary? I don't see how this can be possible without contradicting either ohm's law (I = V/R) or the relationship between the current and voltage of the primary and secondary. (Vp*Ip = Vs*Is) I was under the impression than there was some sort of exception for the primary because of "inductive reactance".

Vp*Ip = Vs*Is applies to the transformer coils (and it is an idealization ie. it assumes that the transformer does not experience losses. Losses occur in several ways: including power losses due to resistance in the primary and secondary coils).

If you put a lightbulb in parallel to the primary coil, you are applying voltage Vp to the bulb, so the current will be less than if you apply the higher secondary voltage Vs. But adding the lightbulb in parallel does not change Vp, Ip or Vs or Is.

If you add a lightbulb in series with the primary circuit it gets rather complicated. You are effectively adding resistance to the primary coil. This will lower the voltage across the primary coil and will, therefore, lower the voltage of the secondary. However, the lightbulb current (primary current) and, therefore the voltage drop across the lightbulb, will fluctuate with the load on the secondary (and the inductive reactance in the primary).

AM