How do you calculate this?

[cdynasty001]

My question is: how do you calculate the power dissipation in watts in an in series circuit with three resistors and three voltage inputs? I would like to know how to calculate it across any of the resistors. I am confused as to how I am supposed to calculate this when there are three voltage inputs, one in between each resistor. I would appreciate some help.

R1 = 2 ohms. R2 = 5 ohms. R3 = 3 ohms.

 

[dfmischler]

It is a simple series DC circuit. That makes it pretty easy because the voltages and resistances add up in a straightforward way, and the current is the same everywhere in the circuit.

Let's assume that the equivalent series resistances of the voltage sources are unimportant. Capacitance and inductance are unimportant because it is DC.

So the total voltage is 30 + 5 + 15 = 50 volts. The total resistance is 5 + 2 + 3 = 10 ohms. So the current is 50 / 10 = 5 Amps. So the voltage drop of R1 is 5 Amps * 2 Ohms = 10 Volts, and the power dissipated is 10 Volts * 5 Amps = 50 Watts. R2 = 5A * 5 ohms = 25 V; 25 V * 5 A = 125 W. R3 = 5A * 3 ohms = 15V; 15V * 5A = 75 W.

 

[Carultch]

My question is: how do you calculate the power dissipation in watts in an in series circuit with three resistors and three voltage inputs? I would like to know how to calculate it across any of the resistors. I am confused as to how I am supposed to calculate this when there are three voltage inputs, one in between each resistor. I would appreciate some help.

R1 = 2 ohms. R2 = 5 ohms. R3 = 3 ohms.

View attachment 15434

Are the 30V and 15V sources intended to be drawn with the opposite polarity as the 5V source? If so, it is incorrect to add 30V+15V+5V to get the full voltage. It really would be 30V + 15V - 5V.

If you group all the resistors together, and group all the voltage sources together, it will be a more intuitive problem. And it is correct to do this in a simple circuit with linear components, because addition is commutative.


Group all the voltage sources, and you see that you get 30V+15V-5V = 40V.
Group all the resistances, and you see that the total resistance is 2 Ohm + 5 Ohm + 3 Ohm = 10 Ohms.


Therefore, the current is Vtotal / Rtotal = 4A.


Now use P=I^2*R to get the power dissipated in each resistor.

 

[dfmischler]

Are the 30V and 15V sources intended to be drawn with the opposite polarity as the 5V source? If so, it is incorrect to add 30V+15V+5V to get the full voltage. It really would be 30V + 15V - 5V.

Yeah, I also just noticed that the polarity of the 5V is reversed. So calculate based on 40 volts instead of 50.

 

[cdynasty001]

Thank you guys. I was stuck on this question for a while.

 

[wwhitney]

In case it is not clear why it is OK lump the resistors and (signed) voltage sources, here is how I would put it:

The circuit is just one loop, so there is only one current present (which could be considered a trivial case of Kirchoff's Current Law).

Kirchoff's Voltage Law around the loop gives you that the sum of the voltage sources and the voltage drops over the resistors will equal 0. Since addition is commutative, as was mentioned, you can reorder the equation as

V1 + V2 + V3 - I R1 - I R2 - I R3 = 0

where the negative sign on the I R terms represents voltage drop. Or for V = V1 + V2 + V3 (with appropriate signs) and R = R1 + R2 + R3, you have

V - I R = 0

Cheers, Wayne

 

[junkhound]

would appreciate some help.

Not trying to be snarky or anything, but am curious what school you are in and your major? Summer school, or is the problem one on a pre-matriculation aptitude/placement test?

The simple type circuit you have shown indicates that some of the school's instructors are sorely lacking in ability to convey basic concepts. T
The accurate response you received from the other members are straightforward and should be how a basic like this should be presented by a school instructor, esp at college level.