How to find power dissipation in combination circuit?

In summary: Assuming this is the correct equation, I know R (34 Ohms) but that's it, right? I don't know current or voltage (the only voltage I know is what's left over). How can I separate that to find what I need for the 34 Ohm resistor?
  • #1
boilerpilot
9
0

Homework Statement



What is the power dissipated in the 34-Ohm resistor?

SCAN0007.jpg


Homework Equations



P=RI^2

The Attempt at a Solution



I have found the total current in the system as well as the current through the resistor, and keep getting 1.88, which is incorrect. Any suggestions on where I am going wrong?
 
Physics news on Phys.org
  • #2
Nobody can find your mistake without seeing your work.

ehild
 
  • #3
Ok, I guess I need to clarify my work. Here is what I have so far...

Rtotal=35.88
I=8/35.88=.222965
P=E2/R
P=82/34
P=1.8823W

The part that confuses me is the combination circuit part, and I am not sure how that plays into the power dissipation of the resistor...
 
  • #4
The voltage is not 8 V across the 34 ohm resistor. Try to find it out.
The parallel resistors can be replaced by their resultant, and this is in series with the 25 ohm resistor. You know the current flowing both series resistors. What is the voltage across the resultant of the parallel ones?

ehild
 
  • #5
ehild said:
The voltage is not 8 V across the 34 ohm resistor. Try to find it out.
The parallel resistors can be replaced by their resultant, and this is in series with the 25 ohm resistor. You know the current flowing both series resistors. What is the voltage across the resultant of the parallel ones?

ehild

I don't really know what you mean. Isn't voltage equal across a parallel circuit? I also don't know what you mean by two series resistors...Is this circuit a series (with the 16 and 25 Ohm) with a parallel circuit (34 Ohm)? Or a parallel (16 and 34) with one in series (25 Ohm)? I have tried to solve this either way, but I can't figure out how much power is dissipated in that 34 Ohm resistor.
 
  • #6
boilerpilot said:
I don't really know what you mean. Isn't voltage equal across a parallel circuit? I also don't know what you mean by two series resistors...Is this circuit a series (with the 16 and 25 Ohm) with a parallel circuit (34 Ohm)? Or a parallel (16 and 34) with one in series (25 Ohm)? I have tried to solve this either way, but I can't figure out how much power is dissipated in that 34 Ohm resistor.

I'd just combine all the resistors (I think you did this at one point), to find out what the current is through the circuit. All that current goes throught the bottom resistor, so you can calculate the voltage across that resistor. The rest of the voltage is across the parallel combination of resistors to the left, and there's an equation for power dissipated in a resistor when it has a voltage across it...
 
  • #7
berkeman said:
I'd just combine all the resistors (I think you did this at one point), to find out what the current is through the circuit. All that current goes throught the bottom resistor, so you can calculate the voltage across that resistor. The rest of the voltage is across the parallel combination of resistors to the left, and there's an equation for power dissipated in a resistor when it has a voltage across it...

The current through the 25 Ohm resistor is .222965A. This means the voltage across the resistor is 5.574V. This leaves 2.426V unaccounted for. How do I know how much power is dissipated from the rightmost resistor?
 
  • #8
boilerpilot said:
The current through the 25 Ohm resistor is .222965A. This means the voltage across the resistor is 5.574V. This leaves 2.426V unaccounted for. How do I know how much power is dissipated from the rightmost resistor?

That leftover voltage is across both of the right-hand resistors. What's the equation for the power dissipated in a resistor based on the voltage across it and the resistance? You can use that same formula to calculate the power dissipated across each of the right side resistors individually.
 
  • #9
berkeman said:
That leftover voltage is across both of the right-hand resistors. What's the equation for the power dissipated in a resistor based on the voltage across it and the resistance? You can use that same formula to calculate the power dissipated across each of the right side resistors individually.

P=RI2=V2/R

Assuming this is the correct equation, I know R (34 Ohms) but that's it, right? I don't know current or voltage (the only voltage I know is what's left over). How can I separate that to find what I need for the 34 Ohm resistor?
 
  • #10
boilerpilot said:
I don't really know what you mean. Isn't voltage equal across a parallel circuit? I also don't know what you mean by two series resistors...Is this circuit a series (with the 16 and 25 Ohm) with a parallel circuit (34 Ohm)? Or a parallel (16 and 34) with one in series (25 Ohm)? I have tried to solve this either way, but I can't figure out how much power is dissipated in that 34 Ohm resistor.

The voltage is equal across parallel connected resistors. How do you know if two resistors are connected in parallel? Which ones in this circuit are connected in parallel?

ehild
 
  • #11
ehild said:
The voltage is equal across parallel connected resistors. How do you know if two resistors are connected in parallel? Which ones in this circuit are connected in parallel?

ehild

It looks to me like the 16 Ohm and 34 Ohm resistors are in parallel. Is that correct? Does this mean that the 16 Ohm and 34 Ohm resistors each have 2.426V across them?
 
  • #12
boilerpilot said:
It looks to me like the 16 Ohm and 34 Ohm resistors are in parallel. Is that correct? Does this mean that the 16 Ohm and 34 Ohm resistors each have 2.426V across them?

Yes.
 
  • #13
berkeman said:
Yes.

So...for the win...

P=V2/R
P=2.4262/34
P=.1731W

Is this correct?
 
  • #14
Yes.

ehild
 
  • #15
ehild said:
Yes.

ehild

Thanks ehild and berkeman for all the help. I sincerely appreciate it.
 

Related to How to find power dissipation in combination circuit?

1. How do I calculate the total power dissipation in a combination circuit?

The total power dissipation in a combination circuit is calculated by adding up the power dissipation of each individual component in the circuit. This can be done by using Ohm's Law (P = I^2 * R), where I is the current flowing through the component and R is the resistance of the component. Once you have calculated the power dissipation for each component, simply add them together to get the total power dissipation.

2. What is the difference between series and parallel circuits in terms of power dissipation?

In a series circuit, the power dissipation is the same for each component as the current flowing through the circuit is constant. Therefore, the total power dissipation in a series circuit is equal to the sum of individual power dissipation. In a parallel circuit, the power dissipation is divided between the different branches of the circuit, so the total power dissipation is less than the sum of individual power dissipation.

3. Can I use the same formula for power dissipation in both AC and DC circuits?

Yes, the formula for power dissipation (P = I^2 * R) can be used for both AC and DC circuits. However, in AC circuits, the current and resistance may vary over time, so the power dissipation will also vary. In DC circuits, the current and resistance are constant, so the power dissipation remains the same.

4. How does the power dissipation affect the temperature of the components in a circuit?

The power dissipation in a circuit is converted into heat, which can increase the temperature of the components. This is why it is important to calculate the power dissipation and make sure it is within the safe operating limits of the components. Excessive heat can damage the components and reduce their lifespan.

5. Can I reduce the power dissipation in a combination circuit?

Yes, there are a few ways to reduce the power dissipation in a combination circuit. One way is to use components with lower resistance, as this will decrease the power dissipation according to Ohm's Law. Another way is to optimize the circuit design, such as using parallel circuits instead of series circuits. Additionally, using more efficient components, such as LEDs instead of incandescent bulbs, can also reduce power dissipation.

Similar threads

Graphing the Power dissipated in a resistor
    • Introductory Physics Homework Help
Replies
1
Views
875
Power dissipated in this circuit with 2 batteries and 3 resistors
    • Introductory Physics Homework Help
Replies
22
Views
2K
Worst-case power consumed by a circuit with MOSFETs
    • Introductory Physics Homework Help
Replies
2
Views
331
Ohm's Law - Finding Source Voltage
    • Introductory Physics Homework Help
Replies
3
Views
593
Energy dissipated in a circuit
    • Introductory Physics Homework Help
Replies
4
Views
4K
Power Resistor Immersed In Water
    • Introductory Physics Homework Help
Replies
4
Views
2K
Power dissipated in a resistor
    • Introductory Physics Homework Help
Replies
10
Views
2K
What happens to the total power dissipated in the circuit ?
    • Introductory Physics Homework Help
Replies
13
Views
3K
Circuit analysis: Six resistors and two batteries
    • Introductory Physics Homework Help
Replies
1
Views
1K
How Does Power Dissipation in One Resistor Relate to Total Power from Batteries?
    • Introductory Physics Homework Help
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top