How to find missing resister with a given voltage value?

I have two circuits, where resister voltage are given, and I need to
find resistor value.


first circuit is -- both are serial circuits

E= 24v, R1 = 2ohm, R2=4 ohm, and R3= 24W (watts).

Find R. How do I do this, Also please explain the method.


Second question is

E = 20V, R1=2V, R2=40ohm, R3=32ohm. Find resistor R1 using voltage
divider rule.

This is what I have done so far for;

formula; v1 = R1*E/R1

2v=R1*20v/R1
2V=20V
=10ohm

I use this method on another serial circuit with all the ohms,
current, voltage, and I didn't get the right answer. There are no
answer at the back of the book, and I don't know if I am right.

 

You have one simple formula relating I to V for R3. I*V=24,
or I=24/V or V=24/I.
You need to come up with a second formula relating I and V
to R value based on the other two resistors and total
voltage. Then combine these two formulas to solve for R3.

The voltage across R3 must be 24 - R1 drop - R2 drop. Each
of these drops is pretty easy to describe in terms of the
common current, which is also pretty easy to describe in
terms of R3.

The divider rule is

fraction of total voltage that is dropped across Rx
= Rx/Rtotal

So, in this case Vfraction = 2/20 = 0.1
=R1/(R1+R2+R3)

 

You can do this with Ohm's Law.

Do you know it ?

Graham

 

In both circuits E=I*(R1+R2+R3) and you have to solve for an unknown R,
which means you must eliminate I from the equation.
In the first circuit you are given the power in R3 is 24W which I^2*R3
so that I=sqrt(24/R3) and the equation becomes:
E=sqrt(24/R3)*(R1+R2+R3), or 24=sqrt(24/R3)*(6+R3), which you can now
solve for R3.
In the second circuit you are given the voltage drop across R1 is
2V=I*R1 so that I=2/R1 and the equation becomes E=2/R1*(R1+R2+R3), or
20=2/R1*(R1+72), which you can now solve for R1.