AP Physics Circuit Question

Question

https://app. gemoo.com/share/image-annotation/649550716401061888?codeId=MpOZa76nzpWON&origin= imageurlgenerator&card= 649550715587366912

(Included spaces in the link so that it would work)

The answer on the key starts by saying, "Using Kirchhoff's loop rule, the voltage across the variable resistor is 5 V." I don't understand why that is the case, and moreover I overall don't know how to do the problem.

Answer 1

First of all, let's handle your question about the 5V.

Let A and B be the two nodes (those big dots) on both sides of the resistor R.

The voltage difference between A and B is simultaneously the voltage difference

across the resistor R as well as the 1000Ω resistor.

(This is a consequence of the assumption of no resistance in the wires,

which implies no voltage drop across wires.)

This is the reason why the voltage across the resistor R is the same as over the 1000Ω resister,

namely the given 5V.

(This could also be derived more formally using Kirchhoff's loop law.)

Once we have the 5V between A and B, this leaves 20V for the 2000Ω resistor.

That means, that the voltage between A and B is four times smaller than the voltage

across the 2000Ω resistor.

That means that the effective resistance, Re, between A and B must be four time smaller,

which is 500Ω.

The effective resistance can be calculated by the formula

1/Re = 1/R + 1/1000

So we have the equations

1/500 = 1/R + 1/1000

Solving this equation gives

R = 1000Ω