Relationship between current and voltage.

[adzbuckland]

Hi guys, this is my first post.

I was just wondering what the exact relationship between current and voltage is.
Transformers work on the basis that P = IV so by increasing the voltage, they decrease
the current and less thermal energy is lost etc...

We can test this out by making V the subject of the equation:

V = P/I
if V = 100J/20A
∴ V = 5V

Here, I have started with a current of 20 amps and a power of 100 joules. The resultant voltage is therefore 5V. Now, if we decrease the current and keep the power the same, we see an increase in voltage, as expected:

V = P/I
if V = 100J/10A
∴ V = 10V

HOWEVER, the equation for voltage seems to contradict this:

V = IR
∴ V = 10A X 10Ω
∴ V = 100V

Here, I have started with a current of 10 amps and a resistance of 10 ohms. The resultant voltage is therefore 100V. Now, if we decrease the current and keep the resistance the same, the voltage decreases:

V = IR
∴ V = 5A X 10Ω
∴ V = 50V

Therefore, when using Power equation, as current decreases, voltage increases. The Voltage equation on the other hand exhibits exactly the opposite effect.

Please can somebody explain why they don't agree with each other (I may just be being thick but I'm only studying GCSE physics... I'm no physicist - by the way sorry for the long post)

Thanks guys.

 

[Drakkith]

I believe that when you use a transformer to step up the voltage, the impedance (resistance) of the secondary side is higher than the primary side. So if you increase the voltage from 10 volts to 100 volts, then the impedance would increase by 10 times as well in order to reduce the current by 10x and keep the power the same. Someone correct me if I'm wrong.

 

[analogdesign]

I'm a little confused about your confusion (sorry if that's confusing ;) )

In the first bit, you hold power constant and see that voltage and current move in opposite directions. This is expected from P = IV.

In the second bit, you hold resistance constant (NOT power) and are surprised that voltage and current move in opposite directions. But since V = IR or R = V/I it follows that holding R constant means V and I must move in opposite directions.

You aren't making an apples-to-apple comparison so the equations do not contradict anything.

Here's another way to see that.

P = IV. But V = IR so P = I^2*R.

Can you see that holding P constant is not the same thing as holding R constant?

 

[analogdesign]

I believe that when you use a transformer to step up the voltage, the impedance (resistance) of the secondary side is higher than the primary side. So if you increase the voltage from 10 volts to 100 volts, then the impedance would increase by 10 times as well in order to reduce the current by 10x and keep the power the same. Someone correct me if I'm wrong.

Yes, one application of a transformer is to change the impedance of a circuit. That's why you can use them to get voltage gain (but not power gain).

 

[cabraham]

Impedance scales with turns ration squared. For a 10:1 ratio, the impedance scales by 100.

 

[adzbuckland]

I'm a little confused about your confusion (sorry if that's confusing ;) )

In the first bit, you hold power constant and see that voltage and current move in opposite directions. This is expected from P = IV.

In the second bit, you hold resistance constant (NOT power) and are surprised that voltage and current move in opposite directions. But since V = IR or R = V/I it follows that holding R constant means V and I must move in opposite directions.

You aren't making an apples-to-apple comparison so the equations do not contradict anything.

Here's another way to see that.

P = IV. But V = IR so P = I^2*R.

Can you see that holding P constant is not the same thing as holding R constant?

Ok thanks for clearing that up haha :)