Is the Calculated Current of 1.3 Amps Correct in a Parallel Circuit Problem?

[zaddyzad]

Homework Statement

Homework Equations

What is the current leaving the cell.

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 ?

 

[Infinitum]

Hi zaddzad!

[strike]Your answer is correct[/strike]. 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...

 

[zaddyzad]

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 ?

 

[Infinitum]

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

[tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}[/tex]

And for a series circuit, the equivalent resistance is,

[tex]R = R_1 + R_2 + ... + R_n[/tex]

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

 

[OmCheeto]

Homework Statement

View attachment 48447

Homework Equations

What is the current leaving the cell.

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.

 

[Infinitum]

I don't get 1.3 amps.

Practicing delta-star circuits currently, makes me think everything is given in ohms!

 

[zaddyzad]

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 ?

 

[zaddyzad]

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?

 

[zaddyzad]

Then I used P=V^2/R to find their ohm values ?

 

[zaddyzad]

How is the correct answer found, and what is it?

 

[Infinitum]

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.

 

[zaddyzad]

What is it then?

 

[zaddyzad]

It's no help having someone tell you your answer is wrong, some guidance would be kindly accepted.

 

[zaddyzad]

Still needing help...

 

[zaddyzad]

help please!

 

[SammyS]

Homework Statement

View attachment 48447

Homework Equations

What is the current leaving the cell.

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 ?

Please give the complete statement of the problem as it was given to you.

 

[Dickfore]

Let the resistors have resistances R1, R2, and R3. As you had written, the current and voltage drop across a resistor is:
[tex]
I = \sqrt{\frac{P}{R}}, \ V = \sqrt{P \, R}
[/tex]

Resistors 2 and 3 are connected in parallel, therefore the voltage drop across them is the same.
[tex]
V_2 = \sqrt{P \, R_2} = V_3 = \sqrt{P \, R_3} \Rightarrow R_2 = R_3
[/tex]
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:
[tex]
I_1 = 2 I_2, \ P = I^{2}_{1} \, R_1 = I^{2}_{2} \, R_2 \Rightarrow 4 R_1 = R_2
[/tex]


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.

 

[OmCheeto]

Here's a hint.

The solution is so easy, I can do it in my head!

 

[OmCheeto]

Practicing delta-star circuits currently, makes me think everything is given in ohms!

I think this is why SammyS just requested verification of the problem.

The graphic doesn't strike me as a typical homework problem.

 

[azizlwl]

The job of the cell is to supply the power.
The total output of the cell must be equal to total output of the load.

For house electric consumption, you're charged by the amount of load total wattage irrespective how you connect your appliances, series or parallel.

 

[ehild]

Azizlwl has right, concentrate to the power data. The power supplied by the source equals to the sum of power dissipated on the resistors.
And you know how the power supplied by the battery is related to the voltage and the current of the battery.

You do not need the resistance values.

ehild

 

[Dickfore]

The job of the cell is to supply the power.
The total output of the cell must be equal to total output of the load.

For house electric consumption, you're charged by the amount of load total wattage irrespective how you connect your appliances, series or parallel.

I can confirm that if you followed the method I had suggested, you would get the same answer.

 

[SammyS]

Homework Statement

Homework Equations

What is the current leaving the cell.

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 ?

In my opinion, this problem is in need of a precise statement.

While it makes sense to assume that the problem is giving the power being dissipated by each resistor, it's also reasonable to assume that the problem is giving the power rating of some ohmic devices, where the power rating is for some standard voltage, such as 120 V.

Assuming that we are given the power actually dissipated by the resistors in this circuit, I get that the current through the power source is 0.15 A . Looking at that answer, I see a quicker way to get it than what I went through. Asking, "What is the total power delivered by the 120V source?", gets the answer immediately. Of course, that's probably the method that OmCheeto used when working the problem in his head.

**************

I see that OP has not been in this discussion for a while. This seems to happen fairly frequently.