Resistance of each IDENTICAL resistor?

[itsagulati]

Homework Statement

The circuit in the drawing contains five identical resistors. The V = 50 V battery delivers 46 W of power to the circuit. (Picture should show up somewhere)

Q: what is the resistance in each resistor

Homework Equations

so far the only things i have thought of is P = IV...which gives me I...and I = V/R

The Attempt at a Solution

I use the above equations to get to a Total resistance of 54.3 ohms...

i thought of several different ways to go about it from there and none of them have worked. dividing by 5 not going to work...
tried multiplying by 4 and dividing by 5 (5 total resistors, but when we add them in parallel we only have 4 being added into the total since two are in series).

anyways...basically i am lost...i've gotten through all 17 problems on the homework and cannot really think this one through...its 3:30 a.m...any help would be GREATLY appreciated!

Thanks!

 

[Astronuc]

P = IV so use I = P/V to get I.

Then one has to find the resistances using V = IR.

So all resistors have same resistance R, but the total resistance is not 5R or 4R.

So one has to find the equivalent resistor noting that R3 and R4 are in series, so the combined resistance is 2R.

Then that 2R is parallel (||) with R2 and R5. So find the equivalent resistance of that combination.

Then that combination is in series with R1, so the total resistance is R1 + R_eq(R2 + R3||R4 + R5), with each resistance being the same R.

 

[Vijay Bhatnagar]

First determine the equivalent resistance of the circuit. R3 and R4 are in series. Hence they add up to (R3 + R4) = 2R (where R is the value of each resistance in ohm). This 2R, R2 and R5 are in parallel. Determine their equivalent value using the relevant formula. Then this equivalent resistance is in series with R1. Add them up to get the equivalent resistance of the circuit (Reqv). Reqv will be in terms of R.

Power consumed = V^2/Reqv . As V is given, Power is given and Reqv is in terms of R, solve to get value of R.

 

[itsagulati]

Thanks!

Ok got it...took a moment to figure out the addition of the fractions in parallel...but sure enough it worked :)

I think the V^2 / R = P was the part that cracked it for me. P = IV wasn't seeming to swing it for me :)

In case anyone else needs it and for someone else to double check that I didn't do this wrong some how:

i did 1/r + 1/2r + 1/r = 5/2r ohms^-1 = 2R/5 ohms + R = 7R / 5

50^2 *5 / 46 = 7R

R = 38.82 Ohms

Thanks again to both of you!