# Finding Resistance

Discussion in 'Homework Help' started by cheyenne, Jan 16, 2008.

1. ### cheyenne Thread Starter Active Member

Dec 6, 2007
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0
The question is: In the circuit attached, the power dissipated in R1 is twice that dissipated in R2 and 4 times that dissipated in R3. Determine the value of R1, R2, R3 and V if I3 is 2A.

Can anyone help me here?
If R1 dissipates 4 times what is dissipated in R3 does this mean the current in R1 is 4 times the current in R3? So it would be 8A. I am on the right lines here? Please help me with this question thanks.

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2. ### recca02 Senior Member

Apr 2, 2007
1,211
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power = square of current times resistance.

and for parallel since voltage is constant
current*resistance will remain same for r3 and r2

use this to calculate total current flowing thru the series resistance.

3. ### cheyenne Thread Starter Active Member

Dec 6, 2007
48
0
Okay so the voltage drop across R2 and R3 are the same.
In R3, P=(I^2)*(R) = 4R

I'm still not making much progress, could you please give me a clue to get me going? Thanks for your help so far.

4. ### recca02 Senior Member

Apr 2, 2007
1,211
0
r2 r3 in parallel
p2/p3=2
here v^2/r = power (more convenient for calculation)
thus r2 is half of r3
i3=2A
hence,
i2=4A
i1=6A
from fig it seems voltage across r1=12v
thus r1=2
power consumed by r1=sq(i)*r1=6*6*2=72
using the data(comparisons of power) in problem find power across other resistance and hence the other resistances.
find equivalent resistance of the circuit...that multiplied by i1 will give total voltage.

5. ### cheyenne Thread Starter Active Member

Dec 6, 2007
48
0
Thanks very much for you help. I just have one more question, is P2/P3 = 2 because the power dissipated in R1 is 2 times what is dissipated in R2 and 4 times what is dissipated in R3, meaning P2 is 2 times P3? If I am wrong can you explain. Thanks

6. ### recca02 Senior Member

Apr 2, 2007
1,211
0
yeah exactly.
do u need an explanation for that?
simple
x:Y:Z=1:2:4