Use Ohm's law and KVL to find the value of R:

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SUMMARY

The discussion focuses on applying Ohm's Law and Kirchhoff's Voltage Law (KVL) to determine the resistance value (R) in a circuit. The calculations demonstrate that with a voltage of 150V and a potential drop of 50V across resistor R, the current through R is 25A, leading to a final resistance value of 2Ω. The participants confirm the correctness of the approach, emphasizing the importance of recognizing voltage drops across parallel resistors to ensure accurate calculations.

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  • Understanding of Ohm's Law (V = IR)
  • Familiarity with Kirchhoff's Voltage Law (KVL)
  • Knowledge of parallel resistor circuits
  • Basic circuit analysis techniques
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Homework Statement



http://img853.imageshack.us/img853/2038/homeworkprob18.jpg

(everything copied above is as given, except 'I3' was supposed to just be 'I', but more simple for me if 'I3')

Use Ohm's law and KVL to find the value of R

Homework Equations


V= IR

KVL, and maybe KCL:

KVL = Summation of potential in a loop = 0

The Attempt at a Solution

150V - 50V - (I1)(20Ω) = 0

100V = (I1)(20Ω)

I1 = 5A

V (on diagram)= (20Ω)(5A)

V = 100V

and I just go straight to saying V = 100V over the 5Ω resistor because it is in parallel with V on the diagram, right?

So, 100V = (5Ω)(I2)

I2 = 20A

I1 + I2 = I3

I3 = 25A

V= I/R ; R = V/I

R = 50V / 25A because 50V was the potential drop across R

R = 2Ω

I didn't see the potential drop across R until I was halfway posting this, but is this correct anyway? It looks weird, and I do not know what the correct answer is.
 
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Your result looks fine.

Yes, you can go right to "100V/5Ω". In fact, you should recognize that the 50V dropped across R means that the parallel pair of resistors must have 100V across it (that's KVL combined with the shared potential drop of parallel branches) so that the total drop equals the supply voltage.

Also, knowing the potential drop across the 5 and 20 Ohm resistors tells you the current through each of them. This should tell you something about the current through R, and knowing both the current and potential drop for R...
 

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