Can anyone help me answer a physics question?

Question

image can be found here:

http://i116.photobucket.com/albums/o36/eniluap99/p18-56alt.gif

how do i go about finding the unknown (X) resistor?

can anyone help me? this is all that is given.

Answer 1

You cannot find the value of the unknown resistor R. It can be anything.

We can use the node-voltage method of analysis to determine R. The node-voltage method sums up the currents into a particular node. Let us take the node above the 30Ω resistor (Call it V2).

The node voltage equation looks like:

(V2 - ΔV)/5 + V2/30 + V2/(40 + R) = 0

You would need two other pieces of information to solve for R. You would have to know either the current through any of the nodes in the system, the voltage ΔV, or the voltage V2.

Let's say you know the current (i) and voltage (v) supplied by the voltage source.

Then the voltage drop across the 5Ω resistor is 5I, so the voltage at the node above the 30Ω resistor is V-5I. The node voltage equation is then:

i + (v-5i)/30 + (v-5i)/(40+R) = 0

And you can simply solve for R, since it would be the only unknown in the equation.

But you definitely need MORE INFORMATION!!!

Answer 2

(R in series with 40 Ω)

(R in series with 40 Ω) in parallel with 30 Ω

(R in series with 40 Ω) in parallel with 30 Ω in series with 5 Ω

R in series with 40 Ω = R +40

(R in series with 40 Ω) in parallel with 30 Ω = (R + 40) * 30 / (R + 40) + 30

= (30R + 120) / R + 70

((R in series with 40 Ω) in parallel with 30 Ω) in series with 5 Ω =

((30R + 120) / R + 70) + 5

<=>

Answer 3

Start with the 40 ohm and R resistors. They are in series so they may be combined by addition.

Rx = 40 + R

You now have the 30 ohm and Rx in parallel and they are combined by summing the inverses:

1/Ry = 1/30 + 1/Rx

1/Ry = 1/30 + 1/[40 + R]

30[40 + R] = [(40 + R) + 30]Ry

30[40 + R] = [70 + R]Ry

Ry = 30[40 + R] / [70 + R]

You now have two resistances in series, Ry and the 5 ohm one. Add these to get the total resistance of the network.

Rt = 5 + Ry

Rt = 5 + 30[40 + R] / [70 + R]

Rt = [5(70 + R) + 30(40 + R)] / [70 + R]

Rt = [350 + 5R + 1200 + 30R] / [70 + R]

Rt = [1550 + 35R] / [70 + R]

Now find R by useing V = I*R (I was not specified but you need a current)

dV = I*Rt = I* [1550 + 35R] / [70 + R]

(dV/I)[70 + R] =  [1550 + 35R]

(dV/I)R  - 35R = 1550 -  70(dV/I)

R[(dV/I) - 35] = 1550 - 70(dV/I)

R = [1550 - 70(dV/I)] / [(dV/I) - 35]

Sort of a check:

suppose R = 0 then what does this say?

dV/I = 1550/70 = Rt

From the original circuit:

Rt = 5 + 1/[1/30 + 1/40]

Rt = 5 + 1/[ 70/1200] = 5 + 1200/70 = 1550/70

So it looks OK for R = 0

Answer 4

Not enough information. You would need to know a current or another voltage or something.

Answer 5

There is no X in the drawing.

But the drawing cannot be solved, there are two many unknowns. There is an unknown resistor "R", and an unknown battery voltage.

.