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

In summary: For the batteries, the current direction is given by the problem statement, part (a).For the resistors, the current direction is given by the voltage drop across the resistor and the resistor polarity. For the central branch R2 the voltage drop is positive, so the current is flowing clockwise. For the lower branch R3 the voltage drop is negative, so the current is flowing counter-clockwise.As you compute the power dissipation for each resistor, keep in mind that you've been given the current direction for each resistor so you know whether you're computing the power delivered or absorbed by the resistor and that should guide you as to whether you use I2R or
  • #1
Color_of_Cyan
386
0

Homework Statement


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.)
28-p-021-alt.gif


(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.

Homework 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 ?

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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  • #2
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?
 
  • #3
gneill said:
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.
 
  • #4
Color_of_Cyan said:
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?
 
  • #5
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) ?
 
  • #6
Color_of_Cyan said:
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.
 
  • #7
gneill said:
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?
 
  • #8
Color_of_Cyan said:
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).
 
  • #9
gneill said:
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?
 
  • #10
Color_of_Cyan said:
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.
 
  • #11
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!
 
  • #12
Color_of_Cyan said:
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!
 
  • #13
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 ANext 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.
 

1. How do you calculate the energy delivered by a battery to resistors in a circuit?

The energy delivered by a battery to resistors in a circuit can be calculated using the formula E = V * I * t, where E is the energy in joules, V is the voltage of the battery in volts, I is the current flowing through the circuit in amperes, and t is the time in seconds.

2. How does the energy delivered to resistors by a battery affect the brightness of a light bulb?

The energy delivered to resistors by a battery directly affects the brightness of a light bulb. A higher energy output from the battery will result in a brighter light bulb, while a lower energy output will result in a dimmer light bulb. This is because the energy is converted into light and heat in the filament of the light bulb, causing it to emit more or less light.

3. What is the relationship between the energy delivered by a battery and the resistance of a circuit?

The energy delivered by a battery is inversely proportional to the resistance of a circuit. This means that as the resistance of a circuit increases, the energy delivered by the battery decreases, and vice versa. This relationship is described by Ohm's Law, which states that V = I * R, where V is the voltage of the battery, I is the current flowing through the circuit, and R is the resistance of the circuit.

4. Can the energy delivered to resistors in a circuit be increased by adding more batteries?

Yes, the energy delivered to resistors in a circuit can be increased by adding more batteries. This is because the total voltage in the circuit will increase, resulting in a higher energy output. However, it is important to keep in mind that the total resistance in the circuit will also increase, which can affect the overall energy delivery.

5. How does the type of battery used affect the energy delivered to resistors in a circuit?

The type of battery used can greatly affect the energy delivered to resistors in a circuit. Different types of batteries have different voltage outputs, which will directly impact the energy delivered. For example, a 9-volt battery will deliver more energy than a 1.5-volt battery. Additionally, the type of battery can also affect the overall lifespan and efficiency of the circuit.

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