Determining the power delivered by a battery -image included

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Discussion Overview

The discussion revolves around determining the power delivered by a battery in a circuit involving multiple resistors. Participants are attempting to solve a homework problem that requires applying Ohm's law and power equations to find the correct power output, while also addressing misunderstandings about the configuration of the resistors.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the power delivered by the battery using the formula Pe= E x Is, arriving at 3450 Watts, but questions the correctness of their approach when the expected answer is 1.56 kW.
  • Another participant suggests checking the resistance to current for voltage calculations using Ohm's law.
  • There is a discussion about the correct way to calculate the current through the resistors, with one participant noting that the 2 and 8 ohm resistors are in series and can be replaced by a single 10 ohm resistor.
  • One participant attempts to replace the 2 and 8 ohm resistors with a 10 ohm resistor and recalculates the current, obtaining 1380 Watts, but still finds it incorrect.
  • Another participant provides a breakdown of the current through various resistors, calculating a total of 26A and arriving at 1560 Watts for the power delivered.
  • There is confusion regarding the series and parallel connections of the resistors, with participants questioning how to combine them correctly.
  • Clarifications are made about the connections of the resistors, particularly the relationship between the 5 ohm resistor and the battery, and the series connection of the 2 and 8 ohm resistors.
  • One participant suggests redrawing the circuit to better understand the connections, indicating that the circuit's presentation may be misleading.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the power delivered by the battery, with some proposing different methods and interpretations of the circuit configuration. The discussion remains unresolved regarding the proper calculations and the configuration of the resistors.

Contextual Notes

Participants highlight potential misunderstandings regarding the arrangement of resistors in series and parallel, as well as the implications of these arrangements on current and voltage calculations. There are also references to specific resistor values and their contributions to the overall circuit behavior, which may depend on the circuit's visual representation.

Josh225
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Homework Statement


See attachment for the question and the picture

Homework Equations


Pe= E x Is
Is = V/R

The Attempt at a Solution


I figured that in order to find the power delivered by the source, I would have to use the equation: Pe= E x Is.
E is already given to me.. 60.
That means I just have to find Is. I used the formula Is= V/R and applied it to each resistor to find the current. Here is what I got:
I1= 12A
I2= 30 A
I3= 7.5 A
I4= 3 A
I5= 5 A
I added these values up to find Is and got 57.5 A. Now I apply the formula Pe= E x Is (Pe= 60 x 57.5) and got 3450 Watts for the power delivered by the battery.

However.. The correct answer is 1.56 kW

Am I using the wrong formula completely? Where did I go wrong?
Thanks!
 

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Use ohms law to check your resistance to current for voltage calculations.
 
Josh225 said:

Homework Statement


See attachment for the question and the picture

Homework Equations


Pe= E x Is
Is = V/R

The Attempt at a Solution


I figured that in order to find the power delivered by the source, I would have to use the equation: Pe= E x Is.
E is already given to me.. 60.
That means I just have to find Is. I used the formula Is= V/R and applied it to each resistor to find the current. Here is what I got:
I1= 12A
I2= 30 A
I3= 7.5 A
I4= 3 A
I5= 5 A
I added these values up to find Is and got 57.5 A. Now I apply the formula Pe= E x Is (Pe= 60 x 57.5) and got 3450 Watts for the power delivered by the battery.

However.. The correct answer is 1.56 kW

Am I using the wrong formula completely? Where did I go wrong?
Thanks!
You are calculating the current through the 2 and 8 ohm resistors as though they were each separately across the 60V battery. Actually the 2 and 8 ohm resistors are in series and the two of them could be replaced by a single 10 ohm resistor across the battery.
 
nsaspook said:
Use ohms law to check your resistance to current for voltage calculations.
I checked the math, and everything checked out. Am I going about the problem in the right way?
 
The Electrician said:
You are calculating the current through the 2 and 8 ohm resistors as though they were each separately across the 60V battery. Actually the 2 and 8 ohm resistors are in series and the two of them could be replaced by a single 10 ohm resistor across the battery.
I tried replacing the 8 and 2 ohms with a 10 ohm resistor. After I added them up, I got Is= 23. I threw that into the formula.. 60 (23) and got 1380 watts.
Still didn't work out.

Is that what you meant to do?
 
Current in:

5 ohm = 12A
10 ohm = 6A
20 ohm = 3A
12 ohm = 5A

12+6+3+5 = 26A

60V*26A = 1560 watts
 
The Electrician said:
Current in:

5 ohm = 12A
10 ohm = 6A
20 ohm = 3A
12 ohm = 5A

12+6+3+5 = 26A

60V*26A = 1560 watts
I see now.. But how do you know to combine the 2 and 8 ohm resistors? Arent the 5 ohm and 2 ohm resistors in series?
 
Last edited:
Perhaps you are thinking that the wire btween the 2 and 8 ohm resistors is connected to the wire between the + terminal of the battery and the top of the 20 ohm resistor. The 2 and 8 ohm resistors are in series only and the wire between them doesn't connect to anything else.

What 1 ohm resistor are you talking about? I don't see any such resistor.
 
The Electrician said:
Perhaps you are thinking that the wire btween the 2 and 8 ohm resistors is connected to the wire between the + terminal of the battery and the top of the 20 ohm resistor. The 2 and 8 ohm resistors are in series only and the wire between them doesn't connect to anything else.

What 1 ohm resistor are you talking about? I don't see any such resistor.
I apologize, I meant the 5 ohm resistor.
 
  • #10
The left end of the 5 ohm resistor is connected to ground; the negative terminal of the battery is also connected to ground. Therefore the 5 ohm resistor is connected across the battery. The 2 and 8 ohm resistor are connected in series across the battery, but with respect to the battery, the 5 and 2 ohm resistors are not connected in series.
 
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  • #11
The Electrician said:
The left end of the 5 ohm resistor is connected to ground; the negative terminal of the battery is also connected to ground. Therefore the 5 ohm resistor is connected across the battery. The 2 and 8 ohm resistor are connected in series across the battery, but with respect to the battery, the 5 and 2 ohm resistors are not connected in series.
Aaaahhhh.. Makes sense now. Thanks a lot!
 
  • #12
This circuit is shown in such a way as to trick you. I usually helps to redraw the circuit. Then you can more clearly see how things are hooked up:

Trick.jpg
 
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  • #13
The Electrician said:
This circuit is shown in such a way as to trick you. I usually helps to redraw the circuit. Then you can more clearly see how things are hooked up:

View attachment 102851
Alrighty! Ill keep that in mind. It's much easier to see that way.
 

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