Solving Electromotive Force Homework: 5000V Power Supply & 20kohm Voltmeter

[caltzx]

Homework Statement

A power supplyused in a lab has an emf of 5000V. However, when a voltmeter of resistance 20kohm is connected to the terminal of the power supply, a reading of only 24V is obtained. Explain this conservation and calculate any necessary quantity.

The Attempt at a Solution

Not really sure where to go with this. I think it has something to do with internal resistance and such but... not really sure. Could someone at least point me in the right direction?

Thank you.

 

[LowlyPion]

Maybe start by drawing a circuit diagram.

You have a box that nominally has 5000V.

But when you effectively put a 20KΩ load across it, it only reads 25V.

So what you apparently have is some kind of voltage divider.

Perhaps you want to consider some internal resistance?

 

[nlightner]

I would first start by understanding the concepts of electromotive force and internal resistance. Electromotive force (EMF) is the energy per unit charge provided by a source, such as a power supply, to drive an electric current. Internal resistance is the resistance within the source itself, which can cause a drop in voltage when current flows through it.

In this case, the power supply has an EMF of 5000V, but when a voltmeter with a resistance of 20kohm is connected to it, the reading obtained is only 24V. This can be explained by the internal resistance of the power supply. When current flows through the power supply, some of the energy is lost due to the internal resistance, resulting in a drop in voltage.

To calculate the internal resistance of the power supply, we can use Ohm's Law (V=IR) where V is the drop in voltage (5000V - 24V = 4976V), I is the current flowing through the power supply, and R is the internal resistance. Rearranging the equation, we get R = V/I.

To find the current, we can use the fact that the voltmeter is connected in parallel to the power supply, so the current flowing through the voltmeter is the same as the current flowing through the power supply. We can use Ohm's Law again (V=IR), where V is the reading obtained from the voltmeter (24V) and R is the resistance of the voltmeter (20kohm). Rearranging the equation, we get I = V/R.

Plugging in the values, we get R = 4976V / (24V / 20kohm) = 10.4 ohm. This is the internal resistance of the power supply.

In summary, the reading of only 24V obtained from the voltmeter is due to the internal resistance of the power supply. By calculating the internal resistance, we can better understand the behavior of the power supply and make accurate measurements in future experiments.