Voltage drop across a certain device

  • Thread starter qwerty321
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The sum of the areas of all the trapezes gives you the integral. In your case, you have to sum the areas of the trapezes in each of the time intervals from t=0 to t=1, t=1 to t=2, ..., t=5 to t=6.In summary, to determine the charge, instantaneous power, and total energy consumed by a device given the voltage drop and current through it, you must integrate the current over time. The instantaneous power is the product of voltage and current at a specific instant, while the total energy is the integral of power over time. To integrate a graph, you can use the trapezoidal rule to calculate the area under the curve in each
  • #1
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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  • #2
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:
 
  • #3
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.
 
  • #4
ok i know that..but how can i integrate a graph?
 
  • #5
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.
 
  • #6
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]
 

1. What is voltage drop across a certain device?

Voltage drop refers to the decrease in voltage that occurs when electricity flows through a device. It is measured in volts and is caused by the resistance of the device.

2. Why is voltage drop important?

Voltage drop is important because it can affect the performance and efficiency of a device. Too much voltage drop can lead to a decrease in power and potential damage to the device.

3. How is voltage drop calculated?

Voltage drop can be calculated by using Ohm's Law, which states that voltage drop is equal to the current flowing through the device multiplied by the resistance of the device.

4. What causes voltage drop in a device?

Voltage drop is caused by the resistance of a device, which can be due to factors such as the material the device is made of, the length of the device, and the temperature of the device.

5. How can voltage drop be reduced?

Voltage drop can be reduced by using materials with lower resistance, shortening the length of the device, and keeping the device at a lower temperature. Additionally, using thicker wires and properly maintaining the device can also help reduce voltage drop.

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