Solving Electric Circuit Current & Power: Peak & Min Values

  • Thread starter bon
  • Start date
  • Tags
    Current
In summary, the conversation discusses the calculation of power dissipated in a resistance in an electric circuit, using the equation P(t) = integral from 0 to t of R Io^2 cos^2wt dt. The peak and minimum values of P are found at sin2wt = 0, giving maximum value of Io^2 R and minimum value of 0. The conversation then moves on to finding the time average of <I V'>, where V' is the voltage across an inductance L. The solution is found to be 0, but the reason for this is not stated.
  • #1

bon

559
0

Homework Statement



Given that the current in an electric circuit is I = Iocoswt and that the current passes through a resistance R

1) give an expression for P(t) power dissipated in the resistance. What are the peak and minimum values of P?

Then it defines the time average over 1 cycle as 1/T (integral from 0 to T) of f(t) dt where T=2pi/w

It asks us to find time average of I and P and illustrate the relevant integrals graphically

Homework Equations





The Attempt at a Solution



Ok so I think P(t) = integral from 0 to t of R Io^2 cos^2wt dt?

Now I am stuck :P How do i find maximum and minimum values of P?

to i differentiate using fundamental theorem to say that P'=RIo^2cos^2 wt?

Then find where this is maximum and minimum?

Also any ideas for the next part? thanks>
 
Physics news on Phys.org
  • #2


P is just I^2*R. You're thinking of energy.
 
  • #3


But I is a function of t, so don't you have to integrate over time? ahh no i guess not.
 
  • #4


Ok great. Thanks for your help...

So could you verify that the power is at max/min where sin2wt = 0...so this implies that t=0,pi/2w, pi/w etc...

so maximum value of power is Io^2 R

minimum is 0?

thanks...
 
  • #5


anyone?
 
  • #6


Ok all this done now :)

Just on the last part of the q..

asks me to work out the time average of <I V' > where V' is the voltage across an inductance L i.e. L dI/dt..

I get the solution to be 0..

why is this?
 

What is electric circuit current?

Electric circuit current refers to the flow of electric charge through a closed circuit. It is measured in amperes (A) and is essential for the operation of electrical devices.

How do you calculate the peak value of electric circuit current?

The peak value of electric circuit current can be calculated by multiplying the root mean square (RMS) value of the current by the square root of 2 (1.414). This will give you the maximum value of the current in the circuit.

What is the significance of peak and minimum values in electric circuits?

Peak and minimum values of electric circuit current are important because they determine the power and efficiency of the circuit. The peak value is the maximum amount of current that can flow through the circuit, while the minimum value is the minimum amount of current needed for the circuit to function properly.

What factors affect the peak value of electric circuit current?

The peak value of electric circuit current is affected by the voltage, resistance, and capacitance of the circuit. Higher voltages and lower resistance will result in a higher peak current value, while a higher capacitance will cause the peak current value to decrease.

How do you calculate the power of an electric circuit?

The power of an electric circuit can be calculated by multiplying the voltage by the current. This will give you the amount of energy being transferred per unit time, measured in watts (W).

Suggested for: Solving Electric Circuit Current & Power: Peak & Min Values

Replies
9
Views
1K
Replies
8
Views
203
Replies
13
Views
689
Replies
3
Views
591
Replies
21
Views
283
Replies
1
Views
788
Replies
8
Views
728
Replies
2
Views
774
Replies
52
Views
2K
Back
Top