Solar Cell calculating the power output

Click For Summary
To determine the power output of a solar cell, connecting a voltmeter and ammeter directly may not yield accurate results due to potential energy buildup in the circuit. Instead, measuring current and voltage across varying load resistances at a constant light intensity provides a clearer picture of power output. The maximum power occurs between short-circuit conditions (maximum current, no power) and open-circuit conditions (maximum voltage, no power). Using a capacitor to store energy complicates the process, as voltage and current will not remain constant during charging. A more straightforward approach is recommended for accurate power measurement.
hickyai5
Messages
2
Reaction score
0

Homework Statement


I am asked to plan a practical to determine the power output of a solar cell, i.e how much light energy is converted toe electrical energy per second. Now I thought(obviously) to just connect voltmeter across it with an ammeter and use PVI, however thinking about it, I don't know if this would work. It might just be my circuit theory but there is nowhere for the energy to go out of the circuit and so won't it just build up so that you have all the electrical energy it has produced as well as what it is producing now. I had another method for it, however, from the mark allocation I think it is too complicated especially as at first this seems like a really easy solution. My more complicated method, was to store all the energy for say a one sec burst of light, on a capacitor and then to discharge it through a resistor with a voltmeter across it, then using V= Vo*e^(-t/RC) to calculate initial voltage across the capacitor and ten use E= o.5CV^2 Can someone just explain why the first method is wrong/right cos its proper fuzzled my way fo thinking of things, btw I'm jst finishing a-level physics to go do it at uni, so i understand most things, but this jst got me thinking


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
use a variable resistance/and measure the current and voltage for a range of load resistances at a constant light intensity. If you short the cell you get maximum current and no power, and if you have an open circuit you get maximum voltage and also no power. The maximum power delivered to the load should be somewhere between that.

Your capacitor idea is problematic because the voltage and current won't be constant when charging it.
 
thanks, very much,
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
35
Views
6K
  • · Replies 3 ·
Replies
3
Views
7K
Replies
9
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
3K