xbuoix said:Homework Statement
Why is pmax=v^2/4r?
Homework Equations
I understand that p=v^2/r using the equations:
p=iv and v=ir
The Attempt at a Solution
My problem relates to when I have to find the pmax. Looking over my professors solutions for the homework, he uses pmax=vt^2/4rt.
gneill said:Hi xbuoix, Welcome to Physics Forums.
Can you post the original problem so that we can see the issue in context? It's hard to say anything meaningful about it without knowing where the 't' comes in or how it comes about.
xbuoix said:Thanks, I'm glad to be on this forum!
Sorry, vt and rt is the Thevenin equation for the voltage and resistance.
The question goes like this:
(There is a circuit.)
Draw the circuit on your solutions and show how you obtain a Thevenin equivalent for everything to the left of the load resistor. Then draw the Thevenin equivalent that you obtain connected to the load resistor. Using the results, what is the maximum power that can be delivered to Rl. What is the value of Wl would result in maximum power?
Finding the Thevenin equation, I found:
VT= 30 V
RT=3kohms
Imax=10mA
RLoad=RT
VMax=VT
At first, I thought that P=VT^2/RT. After I turned in my homework, I looked at the solution my professor posts online, and I see that he used P=VT^2/4RT. Where did the 1/4 come from?
gneill said:Okay, it's clearer now (By the way, you might try using the x2 and x2 icons in the edit panel to create superscripts and subscripts).
The 1/4 comes from the mathematical operations involved in determining the conditions for maximum power delivered to the load resistor. Find the value of RL that results in the maximum power delivered to the load, then use that value of RL to determine that power.
xbuoix said:So this case, the RLoad=RT=3kΩ. I'm still confused on how it came about.
gneill said:Have you studied the math behind finding the maximums and minimums of functions?
xbuoix said:As in finding the double derivative of the function? Yes
Oh.. I get it now..I think.
The double integral of v2 would give 1/4? Shouldn't we have taken the derivative instead?
Edit: I meant 1/R. Why are we taking the maximum of the resistance and not the voltage?
Pmax in V^2/4R is a measure of the maximum power output in an electrical circuit. It is calculated by squaring the voltage (V) and dividing it by four times the resistance (R).
According to Ohm's Law, the power (P) in an electrical circuit is equal to the voltage (V) multiplied by the current (I). Since the current in a circuit is equal to the voltage divided by the resistance (I = V/R), we can substitute this into the power equation to get P = V^2/R. This is then simplified to P = V^2/4R to calculate Pmax.
Pmax in V^2/4R is important in understanding the maximum power that can be delivered to a circuit without exceeding the capacity of the circuit components. It also helps in determining the efficiency of a circuit, as well as the potential for damage or overheating.
Increasing the voltage in a circuit will also increase Pmax, as the power output is directly proportional to the voltage. On the other hand, increasing the resistance will decrease Pmax, as the power output is inversely proportional to the resistance in the circuit.
Yes, it is possible to exceed Pmax in V^2/4R in a circuit, especially if there is a sudden surge of voltage or a decrease in resistance. This can result in overheating and damage to the circuit components. It is important to carefully calculate and monitor Pmax to avoid exceeding the safe limits of a circuit.