Power dissipated from resistance

[baldwindc]

Homework Statement

This was a diagram, so I will describe it as accurately as possible.

[Edit]:
12 Volts go through one 2 Ohm resistor which then proceeds through a set of resistors in parallel. One is 20 Ohms, the other two are 10 Ohms.

How much power is being dissipated by one of the 10 Ohm resistors?

Unknown: Current.

Homework Equations

V = IR
P = V * A
1/R(Parallel-Equivelant) = Sum(1/R_n)

The Attempt at a Solution

Can I calculate the Current after the first resistor using V = I * R ? If so, I know how to calculate the total resistance by using 1/R = Sum of (1/R_n). Would the resistance of one resistor just be 1/R = 1/10 --> R = 10?

THanks for any help

 

[OmCheeto]

Homework Statement

This was a diagram, so I will describe it as accurately as possible.

[Edit]:
12 Volts go through one 2 Ohm resistor which then proceeds through a set of resistors in parallel. One is 20 Ohms, the other two are 10 Ohms.

How much power is being dissipated by one of the 10 Ohm resistors?

Unknown: Current.

Homework Equations

V = IR
P = V * A
1/R(Parallel-Equivelant) = Sum(1/R_n)

The Attempt at a Solution

Can I calculate the Current after the first resistor using V = I * R ?

Yes. But only if you know the values of I & R.

If so, I know how to calculate the total resistance by using 1/R = Sum of (1/R_n). Would the resistance of one resistor just be 1/R = 1/10 --> R = 10?

THanks for any help

First you will have to determine the equivalent resistance of the 3 large resistances in parallel. Once you have that, combine it with the 2 ohm resistor, and you can figure out total circuit resistance. This will yield your total circuit amperage from your equation.

Btw, is this your circuit?

 

[baldwindc]

Yes, that is my circuit. When I first viewed it was completely illegible, so I just deleted it. Could have been a rendering error so thank you for correcting that for me.

You say that I must know the values of I and R. Well by V = IR I should be able to calculate this: 12V = I*(2 Ohms), I = 6A. Is this not the current directly after the first resistor?

Conceptually this doesn't make sense to me. It seems as if current would have to be given in the problem statement.

I calculated the total resistance of the parallel resistors to be 20/5 = 4 Ohms. I also know that one 10 Ohm resistor is 1/4 of the total resistance given in that parallel resistor circuit (Since it accounts for 25% of the total resistance). So shouldn't the current drained from that one resistor be (1/4) * (20/5) = (20/20) = 1.

I don't think that's right. I'm missing something.

 

[OmCheeto]

Yes, that is my circuit. When I first viewed it was completely illegible, so I just deleted it. Could have been a rendering error so thank you for correcting that for me.

You say that I must know the values of I and R. Well by V = IR I should be able to calculate this: 12V = I*(2 Ohms), I = 6A. Is this not the current directly after the first resistor?

That is not correct

Conceptually this doesn't make sense to me. It seems as if current would have to be given in the problem statement.

You can calculate total circuit current by following my procedure.

I calculated the total resistance of the parallel resistors to be 20/5 = 4 Ohms.

That is correct.

I also know that one 10 Ohm resistor is 1/4 of the total resistance given in that parallel resistor circuit (Since it accounts for 25% of the total resistance).

It doesn't quite work that way. As you said, the total resistance of the parallel resisters is 4 ohms. 10 ohms is not 1/4 of 4 ohms.

So shouldn't the current drained from that one resistor be (1/4) * (20/5) = (20/20) = 1.

I don't think that's right. I'm missing something.

You are correct, that's not right.

You really need to find the total circuit current at this point.

 

[aralbrec]

You say that I must know the values of I and R. Well by V = IR I should be able to calculate this: 12V = I*(2 Ohms), I = 6A. Is this not the current directly after the first resistor?

No because that 12V is applied across the 2 Ohm resistor *and* the parallel combination of three resistors.

I'm missing something.

You want to transform your circuit so that it is more easily solvable. This is the purpose of finding an equivalent resistance for series and parallel resistors.

Any parallel set of resistors has the same voltage applied across them. This means the total current through the parallel set can be found by adding V/R of each resistor. If you then look at the voltage applied (V) and total current through the parallel resistors (sum of all V/R), a relationship is found between V and total current through the parallel branch: V/I = 1/(1/R1 + ... + 1/Rn). This looks like Ohm's Law with an effective resistance of 1/(1/R1+...+1/Rn). This means you can *replace* your parallel set of resistors with a single resistance without affecting the rest of the circuit. Further, the current you find in this replacement resistor will be the total current through the parallel resistors. You can determine how that current divides between the resistors from their resistance ratios.

The same can be done with series resistances, with the key insight being the same current that flows through one resistor must also flow through a series resistor in order to get from point A to B.

Your homework problem is then to transform your relatively complicated circuit into a simple one of one resistor connected to a battery to find the circuit current and then work from there by unravelling the circuit transformations you made to compute currents and/or voltages across the original resistances that were made part of the effective resistances in the simplified circuit.