Calculating Wattage Rating in Series/Parallel Circuits

[nyxsilverjk]

Homework Statement

When calculating wattage rating, if you have resistors in both a series and in parallel circuits can you still use P = V²/R to solve for the wattage rating over each resistor if you're given the voltage of the battery or do you have to calculate the voltage drop after each resistor in order to find the next resistor watt rating?

Say I'm given a 24 V battery with a series resistor and two resistors that are in parallel with the series. The series resistor is of 6.9 Ω resistance and the parallel circuits have 1.5 Ω and 8.2 Ω resistance. To find the watt rating of the series resistor the equation it would be P = 24²/6.9 Ω. However when you get to the parallel resistors, would you have the equation P₂ = 24²/1.5Ω or would you have to calculate the voltage drop after the first resistor and use it for the parallel resistor equation?

Homework Equations

The Attempt at a Solution

 

[truesearch]

You must use V^2/R for each resistor. V is not 24V for each resistor

 

[gneill]

Your description of the circuit you intend is not entirely clear, as it's difficult to interpret your uses of series and parallel in a unique way. However, in all cases, the "V" in P = V²/R means the potential drop across the individual resistance.

 

[nyxsilverjk]

Sorry for the bad wording. I guess I was really just asking if I had to calculate the voltage drop after each resistor in order to find the following resistors watt rating?

and would it help to calculate the current? If so do I have to take into account all the resistors or just the first one because current should remain constant throughout the whole circuit?

 

[gneill]

Sorry for the bad wording. I guess I was really just asking if I had to calculate the voltage drop after each resistor in order to find the following resistors watt rating?

No worries; it can sometimes be difficult to put into words a picture that you can see clearly in your mind's eye

You need to find the voltage drop across the individual resistor. Sometimes this will involve a fair amount of analyzing of the given circuit, and finding all the individual voltage drops.

and would it help to calculate the current? If so do I have to take into account all the resistors or just the first one because current should remain constant throughout the whole circuit?

You can certainly use current. In fact, for a given resistance the power is also P = I²R. But again, you'll need to find the current passing through each resistor. This is easy for series connected resistances, but currents divide through parallel paths so that if you have a single resistor followed by a parallel pair, they will not all have the same current flowing through them.

 

[nyxsilverjk]

That's pretty much the problem we're given in that picture you just posted. So the best way to go about this is to find the voltage drop after each resistor and then use that drop to find the next rating?

 

[gneill]

That's pretty much the problem we're given in that picture you just posted. So the best way to go about this is to find the voltage drop after each resistor and then use that drop to find the next rating?

Something in your wording suggests that you're thinking that a 'previous' voltage drop applies to the power consumed by the 'next' resistor in line. This is not the case. The voltage drop you want is the drop across the particular resistor itself.

For the pictured circuit you may employ several methods to find the individual currents or voltages. If the circuit describes your problem, then why not assign the corresponding values to R1, R2, and R3, and make an attempt at determining the parameters?

 

[nyxsilverjk]

So I find the voltage drop at r2. I then use the voltage I got at r2 and use it in the equation P = V^2/ R where the R is equal to the resistance for r2? This then finds the overall power for r2?

 

[gneill]

So I find the voltage drop at r2. I then use the voltage I got at r2 and use it in the equation P = V^2/ R where the R is equal to the resistance for r2? This then finds the overall power for r2?

Yes. Rather than saying "the voltage drop at R2", say, "the voltage drop across R2". Imagine placing a voltmeter on the leads of R2; the voltage you read is then the voltage drop across R2.

 

[nyxsilverjk]

Thank you so much for helping me understand this

 

[nyxsilverjk]

No worries; it can sometimes be difficult to put into words a picture that you can see clearly in your mind's eye

You need to find the voltage drop across the individual resistor. Sometimes this will involve a fair amount of analyzing of the given circuit, and finding all the individual voltage drops.

You can certainly use current. In fact, for a given resistance the power is also P = I²R. But again, you'll need to find the current passing through each resistor. This is easy for series connected resistances, but currents divide through parallel paths so that if you have a single resistor followed by a parallel pair, they will not all have the same current flowing through them.

So how do you figure out how much current goes down either parallel path?

 

[gneill]

So how do you figure out how much current goes down either parallel path?

There are various approaches (which I'm sure you will learn about eventually), but one approach is to first combine the two parallel resistors into a single resistance value (symbolically, R2||R3, that is R2 in parallel with R3). With that 'equivalent resistance' in hand you can then find the voltage drops across R1 and the combined resistance (since the combined resistance is in series with R1). With the voltage across the parallel pair you can use the voltage form for the power equation on the individual resistor values, or use Ohm's law to find the individual currents.

 

[nyxsilverjk]

So let's say the voltage of the battery is 24 V and the resistance at R1 = 2.2 k ohms. The equation would be P of R1 = (24)^2/2200 ohms?

Then when you get to the parallel circuit and go to calculate R2 which equals 1.5 k ohms it would be P of R2 = (24)^2/1500 ohms?

 

[gneill]

So let's say the voltage of the battery is 24 V and the resistance at R1 = 2.2 k ohms. The equation would be P of R1 = (24)^2/2200 ohms?

Then when you get to the parallel circuit and go to calculate R2 which equals 1.5 k ohms it would be P of R2 = (24)^2/1500 ohms?

Nope. The 24V is that of the voltage source, it will NOT be the voltage across the individual resistors. Those you must determine by doing some circuit analysis.

Start by finding the equivalent resistance of the parallel pair of resistors, R2||R3.

 

[nyxsilverjk]

Nope. The 24V is that of the voltage source, it will NOT be the voltage across the individual resistors. Those you must determine by doing some circuit analysis.

Start by finding the equivalent resistance of the parallel pair of resistors, R2||R3.

Ok so I attached a copy of the take home test I've been working on. We are allowed to consult outside sources and work with any other student but for clarity I need you to see the picture of the problem I'm working on it's number 6.

To find the quivalent resistance of the parallel resistors it would be 1/(1500 + 8200 + 12000 ohms) correct? The equivalent resistance I got was 4.6 x 10^-5

 

[gneill]

Ok so I attached a copy of the take home test I've been working on. We are allowed to consult outside sources and work with any other student but for clarity I need you to see the picture of the problem I'm working on it's number 6.

So here's the picture:

To find the quivalent resistance of the parallel resistors it would be 1/(1500 + 8200 + 12000 ohms) correct? The equivalent resistance I got was 4.6 x 10^-5

Not quite. The equivalent resistance of parallel resistances is the reciprocal of the sum of reciprocals:
$$R_p = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}}$$
Once you have that equivalent resistance, you can work out the total series resistance presented to the 24V source, and then the current it will be providing.

 

[nyxsilverjk]

So here's the picture:

Not quite. The equivalent resistance of parallel resistances is the reciprocal of the sum of reciprocals:
$$R_p = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}}$$
Once you have that equivalent resistance, you can work out the total series resistance presented to the 24V source, and then the current it will be providing.

So Rp = 1.15Ω?

Then that number is add to 2.2 and 4.7 to get Rtotal which is equal to 8.1?

then from there you can find the current from I = V/R? which I got to equal 2.96A

 

[gneill]

So Rp = 1.15Ω?

Then that number is add to 2.2 and 4.7 to get Rtotal which is equal to 8.1?

Yes. You should keep a couple of extra decimal places in your intermediate values, and always carry the units, so 8.047 kΩ would be a better value

then from there you can find the current from I = V/R? which I got to equal 2.96A? or do I have to convert to K ohms to ohms?

You need to be careful about the decimal place, so if you specify your resistance in kΩ your result will be in milliamps (mA). If you convert to Ohms first, the result will be in amps. It's probably simpler if you work in base units at this stage, so convert resistances to Ohms for your calculations.

 

[nyxsilverjk]

Yes. You should keep a couple of extra decimal places in your intermediate values, and always carry the units, so 8.047 kΩ would be a better value

You need to be careful about the decimal place, so if you specify your resistance in kΩ your result will be in milliamps (mA). If you convert to Ohms first, the result will be in amps. It's probably simpler if you work in base units at this stage, so convert resistances to Ohms for your calculations.

Ok, I think I had an awesome Ah-hah! moment and I think I can finish the problem. When I do can I give my answers and then you can give me a yes or no to if they're right?

 

[gneill]

Ok, I think I had an awesome Ah-hah! moment and I think I can finish the problem. When I do can I give my answers and then you can give me a yes or no to if they're right?

No promises about yes/no; it might be better to give a hint or two If I'm not online, no doubt someone else will be happy to have a look.

 

[nyxsilverjk]

No promises about yes/no; it might be better to give a hint or two If I'm not online, no doubt someone else will be happy to have a look.

It's ok if you can't tell me. But here's my answers going in order from left to right the last one is the bottom left resistor. 19.28 W , 8.64 W , 1.59 W , 1.08 W , 41.18 W

 

[gneill]

It's ok if you can't tell me. But here's my answers going in order from left to right the last one is the bottom left resistor. 19.28 W , 8.64 W , 1.59 W , 1.08 W , 41.18 W

Your values are all three orders of magnitude too high, so it looks as though you used kΩ as though they were Ω. Remember, a kΩ is 1000Ω. Also, the values are close but are not precise. Did you keep enough digits in intermediate values?

It would help if you would show one or two of your calculations to confirm your methodology.

 

[nyxsilverjk]

Your values are all three orders of magnitude too high, so it looks as though you used kΩ as though they were Ω. Remember, a kΩ is 1000Ω. Also, the values are close but are not precise. Did you keep enough digits in intermediate values?

It would help if you would show one or two of your calculations to confirm your methodology.

I went back and did the entire thing and kept my placing past the thousandths. I also did convert my k ohms and m amps. Then two other students and I compared answers and discussed them, we all came to the same consensus. I'm going to my teacher tomorrow during study hall to confirm some minor questions on syntax.

Thank you so much for helping me with this problem, it's greatly appreciated!