How Can Two Different Resistances Dissipate the Same Power?

[eniluap]

Hi, I am having a bit of trouble solving for an unknown current. Can anyone help me out, thanks.


The resistor R in Fig18.56 dissipates 11 W of power. Determine 2 possible values of R (V= 76 V).

My attempt so far,

1st loop: 35I-35I1=76

2nd loop: I=[(70/30)+(R/30)]I1

I plugged equation (2) into (1): 81.667I1 +1.1667I1R-35I1
=76...(3)
Given that:
I1R =11W/I1

plug that into (3): 46.667I1 +12.8337/I1 -76=0

I've tried using quadratic equation to solve for I1, then plugging into P=I^2R to solve for R, but I still keep getting the wrong answer. Please help. Thank you.

 

[AlephZero]

1st loop: 35I-35I1=76

Are you sure about that? Why do you think both currents flowing through 35 ohms resistance?

 

[eniluap]

Is it 35I-30I1=76? plug (2 )into that then solve with IR. In the end, do I use the quadratic equation to solve for I1? thanks.

 

[AlephZero]

Is it 35I-30I1=76?/

That looks better.

The rest of your first attempt was the right idea.

 

[eniluap]

I attempted the problem again with the new 1st loop equation, however I am still getting the incorrect answer for I1 and R. Can someone help me solve the problem? Thanks.

 

[fizzzzzzzzzzzy]

yea, it's not 35I-30I1. I is supposed to be the current before the currents splits, right? I am guessing you got 35 by adding 30 and 5 together, but when does the current I ever go across the 30 ohm resistor? by the time it reaches the 30 ohm resistor, the current is different cause it split.

also, I am not sure how you got the equation for the second loop, neither I = I2 +I3... or V-RI-RI2-RI3... would result in that.

 

[AlephZero]

The OP was using I for the loop current in the first loop and I2 for the loop current in the second loop I think.

Using that notation my loop equations are

5 I + 30(I-I2) = 76
30(I-I2) = (40+R)I2

Eliminating I gives

1550 I2 = 2280 - 35 R I2

Power = R I2^2 = 11

So multiplying the equation by I2,

1550 I2^2 = 2280 I2 - 385

R = 6.76 or 20.36