Find the energy in a circuit, delivered to resistors, by battery, etc


by Color_of_Cyan
Tags: battery, circuit, delivered, energy, resistors
Color_of_Cyan
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#1
May11-12, 01:30 PM
P: 289
1. The problem statement, all variables and given/known data
The circuit shown in the figure below is connected for 2.10 min. (Assume R1 = 6.00 Ω, R2 = 1.60 Ω, and V = 16.0 V.)


(a) Determine the current in each branch of the circuit. *already did*

(b) Find the energy delivered by each battery.

(c) Find the energy delivered to each resistor.

(d) Identify the type of energy storage transformation that occurs in the operation of the circuit.

(e) Find the total amount of energy transformed into internal energy in the resistors.

2. Relevant equations
*Not even sure*

Summation of current through junction = 0

Summation of potential difference in loop = 0

ΔV = IR

U = ΔV / q ?

ΔU + ΔEint = 0 ?
3. The attempt at a solution
Just need help with parts after A, not sure where to start with regards to them.

I already solved for part A with Kirchoff's rules, the currents are:

I (left wire / branch) = 1.254 A down

I (middle wire / branch) = 0.58 A down

I (right wire / branch) = 1.842 A up

any help is appreciated.
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gneill
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#2
May11-12, 02:20 PM
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What is the formula for the energy provided by a voltage source V with a current I flowing OUT of its positive terminal?

What is the formula for the energy dissipated by a resistance with a current I flowing though it?
Color_of_Cyan
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#3
May11-12, 06:42 PM
P: 289
Quote Quote by gneill View Post
What is the formula for the energy provided by a voltage source V with a current I flowing OUT of its positive terminal?

What is the formula for the energy dissipated by a resistance with a current I flowing though it?
I am honestly not too sure ... all I can think of is P = I2R

and P = IΔV

But I can't figure out how it relates to work, and how to apply it here.

gneill
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#4
May11-12, 09:38 PM
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Find the energy in a circuit, delivered to resistors, by battery, etc


Quote Quote by Color_of_Cyan View Post
I am honestly not too sure ... all I can think of is P = I2R

and P = IΔV
Yes, those formulas apply
But I can't figure out how it relates to work, and how to apply it here.
The problem isn't about work, per se, it's about energy either delivered or dissipated. How is power related to energy? What's the specified time interval?
Color_of_Cyan
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#5
May11-12, 10:48 PM
P: 289
I am not even sure how the current / which current applies to what battery... I think it is just for the same wire / branch though (ie only 0.58A applying to just the 4V battery)


I take it though that you calculate power and then multiply by the time? Is this right?

Also forgot P = (ΔV)2 / R

For the middle branch would I simply plug in

P = (4V)2 / 6 ohms

or P = (0.58A)2(6 ohms)

and then multiply either by the time (126 seconds in this case) ?
gneill
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#6
May11-12, 10:55 PM
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Quote Quote by Color_of_Cyan View Post
I am not even sure how the current / which current applies to what battery... I think it is just for the same wire / branch though (ie only 0.58A applying to just the 4V battery)


I take it though that you calculate power and then multiply by the time? Is this right?

Also forgot P = (ΔV)2 / R

For the middle branch would I simply plug in

P = (4V)2 / 6 ohms

or P = (0.58A)2(6 ohms)

and then multiply either by the time (126 seconds in this case) ?
The formula P = V2/R applies when V is the change in potential (voltage) across the resistor R. The 4V you mention is the voltage of the battery in the branch with the 6 Ohms, but it isn't the voltage across the 6 Ohms. However, you DO have the current in the branch and that does flow through the 6 Ohms, so you can use the I2R formula for the resistance.

And yes, when you multiply power (J/s) by time (s) you end up with total energy delivered or dissipated over that time.
Color_of_Cyan
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#7
May11-12, 11:31 PM
P: 289
Quote Quote by gneill View Post
The formula P = V2/R applies when V is the change in potential (voltage) across the resistor R. The 4V you mention is the voltage of the battery in the branch with the 6 Ohms, but it isn't the voltage across the 6 Ohms. However, you DO have the current in the branch and that does flow through the 6 Ohms, so you can use the I2R formula for the resistance.

And yes, when you multiply power (J/s) by time (s) you end up with total energy delivered or dissipated over that time.
Weird... says it is wrong.

Did (0.58A)^2 multiplied by 6 ohms and then multiplied by 126 seconds.

Do I have to account for the current in the left resistor as well or anything?

Or do I use the P = IΔV formula instead for the energy delivered by the battery, and then multiply by time again?
gneill
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#8
May12-12, 06:05 AM
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Quote Quote by Color_of_Cyan View Post
Weird... says it is wrong.

Did (0.58A)^2 multiplied by 6 ohms and then multiplied by 126 seconds.

Do I have to account for the current in the left resistor as well or anything?

Or do I use the P = IΔV formula instead for the energy delivered by the battery, and then multiply by time again?
Which part of the question are you answering? R1 is 6 Ohms, but its current is not 0.58A. The central branch has two resistors so you'll need to calculated the power dissipation for each of them separately. For the batteries you will apply the other formula, P = IΔV and pay attention to the current direction; if current is flowing OUT of a battery's positive terminal then it is delivering energy to the circuit; if current is flowing INTO the battery's positive terminal then it is absorbing energy (it is recharging).
Color_of_Cyan
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#9
May12-12, 03:24 PM
P: 289
Quote Quote by gneill View Post
Which part of the question are you answering? R1 is 6 Ohms, but its current is not 0.58A. The central branch has two resistors so you'll need to calculated the power dissipation for each of them separately. For the batteries you will apply the other formula, P = IΔV and pay attention to the current direction; if current is flowing OUT of a battery's positive terminal then it is delivering energy to the circuit; if current is flowing INTO the battery's positive terminal then it is absorbing energy (it is recharging).
Thank you so much, I got it correct. I meant 6 ohms was for the middle branch also, to use for the current through the battery.



How about for the middle resistors individually?

You said P = I2R can be used for individual resistors, so for the middle resistors can I just use the current 0.58A and then each resistance separately for R and then multiply by time?

Is the current through each resistor in the middle branch negative like in the battery, since the current is flowing toward the battery? Or do I have to use potential?
gneill
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#10
May12-12, 06:02 PM
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Quote Quote by Color_of_Cyan View Post
Thank you so much, I got it correct. I meant 6 ohms was for the middle branch also, to use for the current through the battery.



How about for the middle resistors individually?

You said P = I2R can be used for individual resistors, so for the middle resistors can I just use the current 0.58A and then each resistance separately for R and then multiply by time?
Yes, that's the correct approach.
Is the current through each resistor in the middle branch negative like in the battery, since the current is flowing toward the battery? Or do I have to use potential?
Resistors are "passive" components -- they don't care which direction the current is flowing, they will dissipate the same energy regardless.
Color_of_Cyan
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#11
May12-12, 10:03 PM
P: 289
WebAssign is just weird....

says "You appear to be calculating this correctly using an incorrect value for the current through the 4-V battery" yet says the magnitude of the current downwards in the middle is correct (0.58A). Or is the current actually different from that of the branch when using P = (I^2)(R) ? Because as far as I remember, the current through a single wire with anything connected in series (doesn't matter if they are batteries, resistors, or capacitors) is the same value.


Got 2 of the individual resistors right doing what you said though. All of the energy is also transformed into internal energy so you just add them in the end.


You helped me a lot anyway, thanks gneill!
gneill
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#12
May13-12, 07:57 AM
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Quote Quote by Color_of_Cyan View Post
WebAssign is just weird....

says "You appear to be calculating this correctly using an incorrect value for the current through the 4-V battery" yet says the magnitude of the current downwards in the middle is correct (0.58A). Or is the current actually different from that of the branch when using P = (I^2)(R) ? Because as far as I remember, the current through a single wire with anything connected in series (doesn't matter if they are batteries, resistors, or capacitors) is the same value.
You are correct that the current in series connected components is the same.

Two things occur to me. One, the program may be being picky about accuracy in this case. You are using 0.58 A for the branch current, but should keep a couple more decimal places for purposes of calculation and round only the answers to the required significant figures.
Two, did you submit the value with the correct sign?
Got 2 of the individual resistors right doing what you said though. All of the energy is also transformed into internal energy so you just add them in the end.


You helped me a lot anyway, thanks gneill!
Color_of_Cyan
Color_of_Cyan is offline
#13
May14-12, 01:33 AM
P: 289
Yeah that's what I was thinking to be honest.


I got all of them right now... had to look at the currents again. The current in the middle was actually 0.5877 A not just 0.58 A


Next time I won't round off so much, I just did it because I was in a hurry and didn't feel like writing down that many more digits / numbers.

Again, thanks for the help.


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