Basic Question on Power Calculations

[abhinand rd]

Hello
I have a doubt in the DC power calculations.
If i have a supply with 48V and delivers 5A no matter what load.
And i connect 5 ohm load to this.
if i use Power calculations like this:
V*I=48*5=240W
I*I*R=5*5*5=125W
(V*V)/R=48*48/5=460W.

Why there is difference in the power using these formulas?

 

[KerimF]

You are confused because, in the real world, there is NO power supply that can deliver a constant voltage and a constant current to ANY load.
Let us take two clear examples:

(1) Let us suppose the load is a short circuit (almost zero ohm). What do you expect the voltage of the power supply output would be?

(2) And if the load is an open circuit, is it possible for the power supply providing any current?

Kerim

 

[abhinand rd]

Yes i am confused, and the two examples really didnt take out the confusion.
in first case the current would shoot up
and in second case there wouldnt be any current. but what about power in these scenarios?
considering in real world limitations how do we calculate power? do these formulas have any condition to exist(Except for zero voltage and zero Current)?

 

[KerimF]

in first case the current would shoot up

Actually the voltage output becomes close to 0 V hevce the power is also close to zero Watt (V*I).

and in second case there wouldnt be any current.

As you said I=0, therefore P = 48V * 0A = 0 Watt

In between we have three conditions:
(1) R_load = Vmax/Imax
(2) R_load > Vmax/Imax
(3) R_load < Vmax/Imax
where, from your example:
Vmax = 48 V but let us make it 50 V instead for simplicity. ok?
Imax = 5 A

In case (1):
R_load = 50 / 5 = 10 Ohm
So P = Vmax*Vmax/R_load = 50*50/10 = 250 W
also P = Imax*Imax*R_load = 5*5*10 = 250 W
also P = Vmax*Imax = 50*5 = 250 W

In case (2):
Since R_load is relatively of a high resistance, it doesn't allow a 5A current to pass thru it by a 50V only.
Actually:
I_load = Vmax/R_load
So if R_load is 25 Ohm (> 10 Ohm)
I_load = 50 / 25 = 2 A (not 5 A)
So P = Vmax*Vmax/R_load = 50*50/25 = 100 W
also P = I_load*I_load*R_load = 2*2*25 = 100 W
also P = Vmax*I_load = 50*2 = 100 W

In case (3):
Since R_load is relatively of a low resistance, it needs more current than 5A if 50V is applied on it.
Since I_max of the supply cannot exceed 5A, the voltage on the load HAS to decrease to:
V_load = Imax*R_load (not Vmax anymore)
So if R_load is 2 Ohm (< 10 Ohm)
V_load = 5*2 = 10 V (not 50 V)
So P = V_load*V_load/R_load = 10*10/2 = 50 W
also P = Imax*Imax*R_load = 5*5*2 = 50 W
also P = V_load*Imax = 10*5 = 50 W

Sorry I can't give you more than this

 

[schmitt trigger]

Kerim;
I congratulate you for giving such a well explained and thorough reply...really professional.

 

[KerimF]

:smile: Thank you, schmitt trigger

@abhinand:
For instance here is a riddle about the voltage supply:
You will learn later that if a voltage source (V) has an internal resistance (Rs), the maximum power that the power supply can deliver to an external load (R_load) occurs when R_load = Rs. In this case, the voltage on the load (V_load) is V/2 and:
Pmax = V_load*V_load/R_load
Pmax = (V/2 * V/2) / R_load = V*V//(4*R_load)
So far so good.
Let us assume Rs is close to zero ohm, therefore at maximum power R_load = Rs, hence also R_load is close to zero Ohm
Replacing 0 ohm in the previous Pmax expression:
Pmax = (V*V) / (4*0) = a value going to infinity.

On the other side, when we short cicuit the output of an ideal voltage source (Rs=0), we say that the delivered power is close to zero, not infinity as the previous analysis says Do you know why?

 

[abhinand rd]

thank you Kerim for a thorough explanation.
and no i dont know why power is zero in one case and infinity in another.
can u please explain?

 

[KerimF]

why power is zero in one case and infinity in another

Perhaps I had calling it a paradox

In theory it is infinity.

But in practice, when we short the output of a voltage source (R_load close to zero Ohm), one of three happens:

(1) The voltage supply blows
(2) The internal current limiter keeps the output current at Imax, say 5A. So if the shorting wire is 0.001 Ohm, P_load= 5*5*0.001 = 0.025 W which is relatively close to zero.
(3) The internal short-circuit protection shuts down the output completely hence reduces the output voltage to zero volt.