Total charge lifted by the charge escalator

In summary, the total charge lifted by the "charge escalator" in this problem is 1080 C. The calculation was found by first finding the average voltage of 1.5 V over 2 hours, which is equivalent to 0.15 V over 7200 seconds. Using Ohm's law, we can then find the average current to be 0.15/5 = 0.03 A. To find the total charge, we use the definition of current, which is dQ/dt = I, and integrate both sides to get Q = I x t. Plugging in our values, we get Q = (0.03 A)(7200 seconds) = 1080 C.
  • #1
ozma
4
0
total charge lifted by the "charge escalator"

Homework Statement



A 1.5 V flashlight battery is connected to a wire with a resistance of 5.00 ohms. The figure shows the battery's potential difference as a function of time. The slope is linear going from (0, 1.5 V) to (2 hrs, 0).


Homework Equations



I = [tex]\Delta[/tex]V / R
I = dQ / dt


The Attempt at a Solution



I = 1.5 V / 5 ohms = 0.3 A
0.3 A = dQ / (2 hrs x 60 x 60) --> dQ = 2160 C

This is not the correct answer, and I don't understand why. Thanks for your help.
 
Physics news on Phys.org
  • #2


ozma said:

Homework Statement



A 1.5 V flashlight battery is connected to a wire with a resistance of 5.00 ohms. The figure shows the battery's potential difference as a function of time. The slope is linear going from (0, 1.5 V) to (2 hrs, 0).


Homework Equations



I = [tex]\Delta[/tex]V / R
I = dQ / dt


The Attempt at a Solution



I = 1.5 V / 5 ohms = 0.3 A
0.3 A = dQ / (2 hrs x 60 x 60) --> dQ = 2160 C

This is not the correct answer, and I don't understand why. Thanks for your help.
When you solve the problem like that, you are assuming the current was .3A for the entire 2 hours, but as the graph shows, the voltage and therefore current steadily increase with time.

Set the problem up more like this:
[tex] \frac{V(t)}{R}=\frac{dQ}{dt}[/tex]
You can solve it with one integration. If you don't know integration, you can solve it the way you did and use the average voltage.
 
  • #3


xcvxcvvc said:
When you solve the problem like that, you are assuming the current was .3A for the entire 2 hours, but as the graph shows, the voltage and therefore current steadily increase with time.

Set the problem up more like this:
[tex] \frac{V(t)}{R}=\frac{dQ}{dt}[/tex]
You can solve it with one integration. If you don't know integration, you can solve it the way you did and use the average voltage.

So I tried this:

average of the voltage (1.5 V) over 2 hours = 1.5 / (2 x 60 x 60) = 2.1 x 10^-4

average I = average V / R = 2.1 x 10^-4 / 5 = 4.2 x 10^-5

average I = dQ / dt --> 4.2 x 10^-5 = dQ / (60 x 2 x 2) --> dQ = 0.01

but this answer is still incorrect. Any ideas? Thanks.
 
  • #4


ozma said:
So I tried this:

average of the voltage (1.5 V) over 2 hours = 1.5 / (2 x 60 x 60) = 2.1 x 10^-4

average I = average V / R = 2.1 x 10^-4 / 5 = 4.2 x 10^-5

average I = dQ / dt --> 4.2 x 10^-5 = dQ / (60 x 2 x 2) --> dQ = 0.01

but this answer is still incorrect. Any ideas? Thanks.

The average value of that function is 1.5/2 = .75 volts.

Edit: You should probably solve it the calculus way since you've been told about the derivative definition of current:
[tex] \frac{V(t)}{R}=\frac{dQ}{dt}[/tex]
First we define V(t) as a function of time. It's of the form mx+b since it's linear, and by looking at the graph, you can calculate m and see that b = 0.

multiply both sides by the differential time:
[tex] \frac{mt*dt}{R}=dQ[/tex]
Integrate both sides:
[tex]\int_{t_i}^{t_F}\frac{mt}{R}\, dt=\int_{0}^{Q_{total}} dQ[/tex]
[tex]Q_{total}=\int_{t_i}^{t_F}\frac{mt}{R}\, dt[/tex]
[tex]Q_{total}=\frac{mt^2}{2R}|_{t_i}^{t_F}[/tex]
 
Last edited:
  • #5


xcvxcvvc said:
The average value of that function is 1.5/2 = .75 volts.

Edit: You should probably solve it the calculus way since you've been told about the derivative definition of current:
[tex] \frac{V(t)}{R}=\frac{dQ}{dt}[/tex]
First we define V(t) as a function of time. It's of the form mx+b since it's linear, and by looking at the graph, you can calculate m and see that b = 0.

multiply both sides by the differential time:
[tex] \frac{mt*dt}{R}=dQ[/tex]
Integrate both sides:
[tex]\int_{t_i}^{t_F}\frac{mt}{R}\, dx=\int_{0}^{Q_{total}} dQ[/tex]
[tex]Q_{total}=\int_{t_i}^{t_F}\frac{mt}{R}\, dx[/tex]
[tex]Q_{total}=\frac{mt^2}{2R}|_{t_i}^{t_F}[/tex]

I don't understand why the average V would be 1.5 / 2. Shouldn't I divide by the time in seconds? 2 hours = 2 x 60 x 60 seconds, that's why I used this number as the denominator.

[tex]Q_{total}=\frac{mt^2}{2R}|_{t_i}^{t_F}[/tex]

What numbers do I put in this? Is m the slope of the graph? Would that be 1.5 / 2 or 1.5 / (2 x 60 x 60)? And what am I using for t, 2 or the time in seconds?

If I use t = 2, then I have Q = ((1.5/2) x 2^2) / (2 x 5) = 0.3 C
 
  • #6


ozma said:
I don't understand why the average V would be 1.5 / 2. Shouldn't I divide by the time in seconds? 2 hours = 2 x 60 x 60 seconds, that's why I used this number as the denominator.

[tex]Q_{total}=\frac{mt^2}{2R}|_{t_i}^{t_F}[/tex]

What numbers do I put in this? Is m the slope of the graph? Would that be 1.5 / 2 or 1.5 / (2 x 60 x 60)? And what am I using for t, 2 or the time in seconds?

If I use t = 2, then I have Q = ((1.5/2) x 2^2) / (2 x 5) = 0.3 C

You wouldn't divide by the time to find average voltage. An easy way to check this is to see the units would be volts per second for average voltage... We want volts. The function's average is the middle value of the linear function, because the higher parts past the halfpoint are exactly balanced by the lower parts before it.

As for your second question, if you are integrating, you can use time in hours if the slope is in voltage per hour or you can use seconds if the slope is in voltage per second. It's up to how you find your slope. Slope WOULD be V/s unlike average voltage. m = 1.5/(2*3600) for volts per second.

However, when you're using average voltage, you need to use seconds and cannot use hours or anything due to the definition of current (which is found by V/R). 1.5/2 * 2 * 3600/5 = total charge.
 
  • #7


i did average V / R = (1.5 / 2) / 5 = 0.15
I = dQ / dt --> 0.15 = dQ / (2 x 60 x 60) --> dQ = 1080 C
and it's correct! finally! thanks for your help.
 

1. What is the charge escalator?

The charge escalator is a mechanism that increases the price of goods or services over time to account for inflation and other factors.

2. How does the charge escalator work?

The charge escalator works by regularly adjusting the price of a good or service based on a predetermined rate. This rate is often tied to the rate of inflation, but can also be influenced by other factors such as market demand and supply.

3. How is the total charge lifted by the charge escalator calculated?

The total charge lifted by the charge escalator is calculated by multiplying the original price of the good or service by the rate of the charge escalator over a certain period of time. For example, if a good originally costs $100 and the charge escalator rate is 2% per year, the total charge lifted after 5 years would be $110.40 ($100 x 1.02^5).

4. What are the benefits of using a charge escalator?

The main benefit of using a charge escalator is that it allows businesses to maintain their profit margins over time despite inflation and other economic factors. It also helps to ensure pricing consistency and predictability for both businesses and consumers.

5. Are there any potential drawbacks to using a charge escalator?

One potential drawback of using a charge escalator is that it can lead to higher prices for consumers over time. It can also be difficult to predict the impact of other economic factors on the rate of the charge escalator, which may result in unexpected price increases.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
709
  • Introductory Physics Homework Help
Replies
2
Views
754
  • Introductory Physics Homework Help
Replies
5
Views
945
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
877
  • Introductory Physics Homework Help
Replies
1
Views
757
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
665
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top