## Circuit question, watts!

1. The problem statement, all variables and given/known data

2. Relevant equations

What is the current leaving the cell.

3. The attempt at a solution

Since all the resistors have the same drop in watts, I assumed both "sections" will lose drop 6V. So each side of the the parallel section drops 6V and the single resistor also drops 6V.

Then I tried to find the resistance in ohms of each one. P=V^2/R... 60^2/60=60.
So I assumed every resistor is 60ohms. I then went to find the total resistance of the system (60^-1+60^-1)^-1+60. Rt=90.. Then I used V=IR 120/90 to find current. I end up getting 1.3Amps is this correct ?
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 Hi zaddzad! Your answer is correct. Though, it would be much easier to answer this question using the concept of equivalent resistances(instead of P). You can easily find the equivalent resistance due to the parallel connection, and then due to the series, for total resistance in the circuit. Edit : I just realized the given terms is watts not ohms...
 Hm, don't think I'v ever heard of this concept. I have a circuits test tomorrow, would you mind showing me how it's done ?

## Circuit question, watts!

Basically, when you have a parallel circuit, the equivalent resistance R is given as,

$$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + ....... + \frac{1}{R_n}$$

And for a series circuit, the equivalent resistance is,

$$R = R_1 + R_2 + ......... + R_n$$

Take a look here for a deeper understanding : http://en.wikipedia.org/wiki/Series_...allel_circuits

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 Quote by zaddyzad 1. The problem statement, all variables and given/known data Attachment 48447 2. Relevant equations What is the current leaving the cell. 3. The attempt at a solution Since all the resistors have the same drop in watts, I assumed both "sections" will lose drop 6V. So each side of the the parallel section drops 6V and the single resistor also drops 6V. Then I tried to find the resistance in ohms of each one. P=V^2/R... 60^2/60=60. So I assumed every resistor is 60ohms. I then went to find the total resistance of the system (60^-1+60^-1)^-1+60. Rt=90.. Then I used V=IR 120/90 to find current. I end up getting 1.3Amps is this correct ?
I don't get 1.3 amps.

 Quote by OmCheeto I don't get 1.3 amps.
Practicing delta-star circuits currently, makes me think everything is given in ohms!
 Thats what I did with the P=V^2/R I found the resistance of each resistor. Then when I found the resistance was 60, I did the total resistance of the circuit. For the parallel part I did (60^-1=60^-1)^-1 = 30 resistance there, and for the series I just added 60. The total resistance of the circut is 90 no ?
 I assumed each section of the circuit dropped 60 of the 120 volts because they are = in power drops. So wouldn't the top series resistor drop 60 volts, and the parallel resistors as a whole drop 60 volts?
 Then I used P=V^2/R to find their ohm values ?
 How is the correct answer found, and what is it?

 Thats what I did with the P=V^2/R I found the resistance of each resistor. Then when I found the resistance was 60, I did the total resistance of the circuit. For the parallel part I did (60^-1=60^-1)^-1 = 30 resistance there, and for the series I just added 60. The total resistance of the circut is 90 no ?
The resistance is not 60 ohms for the resistors.
 What is it then?
 It's no help having someone tell you your answer is wrong, some guidance would be kindly accepted.
 Still needing help...
 Let the resistors have resistances R1, R2, and R3. As you had written, the current and voltage drop across a resistor is: $$I = \sqrt{\frac{P}{R}}, \ V = \sqrt{P \, R}$$ Resistors 2 and 3 are connected in parallel, therefore the voltage drop across them is the same. $$V_2 = \sqrt{P \, R_2} = V_3 = \sqrt{P \, R_3} \Rightarrow R_2 = R_3$$ This means that the current passing through each of them is the same. According to 1st Kirchhoff rule, the current through resistor 1, is twice as big. But, the power delivered is the same. Therefore: $$I_1 = 2 I_2, \ P = I^{2}_{1} \, R_1 = I^{2}_{2} \, R_2 \Rightarrow 4 R_1 = R_2$$ We have found the relative resistances of the three resistors. What is the equivalent resistance of the three resistors in terms of R1? What is the current flowing through resistor 1, in terms of V = 120 V, and R1? What is the power delivered on resistor 1, in terms of V, and R1? You know that this power is 60 W. Find R1 from here. Go one step backwards and find the current through resistor 1. This is the same current flowing out of the source.