Voltage drop across a certain device

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Discussion Overview

The discussion revolves around calculating the charge, instantaneous power, and total energy consumed by a device based on provided voltage and current graphs over a time interval from 0 to 6 seconds. Participants seek guidance on how to approach these calculations without requiring direct solutions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant requests help in determining charge, instantaneous power, and total energy from voltage and current graphs.
  • Another participant emphasizes the relationship between current and charge, suggesting that charge can be derived from current over time.
  • It is noted that instantaneous power can be calculated as the product of voltage and current at any given moment.
  • A participant expresses confusion about how to integrate a graph to find these values.
  • Responses clarify that integration can be understood as finding the area under the graph, with one participant suggesting the trapezoidal rule for discrete values.

Areas of Agreement / Disagreement

Participants generally agree on the relationships between charge, power, and energy, but there is uncertainty regarding the method of integrating graphs, particularly for discrete data points.

Contextual Notes

Participants discuss integration methods without resolving the specifics of applying these methods to the provided graphs. There is a lack of consensus on the best approach to integrate the data presented.

qwerty321
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the voltage drop across a certain device and the current through it in the direction of the voltage rop are shown in figure below..Determine:

the charge q though the device at the end of each 1 s interval from t=0 to t=6s.

The instantaneous power during the aforementioned intervals.

the total energy consumed by the device.

i have a figure where it shows the graph of v(y axis) and time(x axis)

and another one showing the graph of i(y axis) and time(x axis)

I am not able to solve this soo please help me..

i know that i must show that i have tried to solve this..i do not want the spolution..juste a guide

thanks really for the help
 
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Welcome to PF!

qwerty321 said:
i have a figure where it shows the graph of v(y axis) and time(x axis)

and another one showing the graph of i(y axis) and time(x axis)

Hi qwerty321! Welcome to PF! :smile:

(always use capitals for V and I :wink:)

You must learn your electric units …

current = charge per time, so charge = … ?

voltage = energy per charge, so energy = … ?

power = current times voltage …

and integrate over t. :smile:
 
qwerty321 said:
the voltage drop across a certain device and the current through it in the direction of the voltage rop are shown in figure below..Determine:

the charge q though the device at the end of each 1 s interval from t=0 to t=6s.

The instantaneous power during the aforementioned intervals.

the total energy consumed by the device.

i have a figure where it shows the graph of v(y axis) and time(x axis)

and another one showing the graph of i(y axis) and time(x axis)

I am not able to solve this soo please help me..

i know that i must show that i have tried to solve this..i do not want the spolution..juste a guide

thanks really for the help

The charge is the integral of the current,
The instantaneous power is the product of voltage and current in the instant considered.
The energy is the integral of the power.
 
ok i know that..but how can i integrate a graph?
 
qwerty321 said:
ok i know that..but how can i integrate a graph?

An integral is just the area under a graph (okay, the area between the graph and the x-axis, positive or negative depending on which side it happens to be on). Antiderivatives just happen to be an easier(?) way of obtaining this number than plotting the function and counting graphing paper squares. Or attempting to fit data to a curve, and then finding the antiderivative of that.
 
qwerty321 said:
ok i know that..but how can i integrate a graph?

Since you have only discrete values in your graph, you can use the trapezoidal rule:
Let f(t) the value of the function at time t and f(t+T) the value at time t + T.
So, you have a trapeze with bases f(t) and f(t+T) and height T. The area of the trapeze is:
[tex]A = \frac{f(t) + f(t+T)}{2T}[/tex]
 

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