Asked by Mike

A certain 9-V battery measures 9.0 V with no load connected to it, but only 8.5 V when connected to a radio which draws 150 mA of current.

What is the internal resistance of the battery?
Answered by bobpursley
.150*internalresistance=9-8.5

Answered by Mike
I see you're basically finding the voltage drop that occurs as a result of that internal resistance.

Now they ask, in the second part .... What would be the terminal voltage for a 50-W load?
Answered by Mike
Oops. question should say 50 ohm load NOT 50-W load. Sorry.


Terminal Voltage as I understand it would be the Vt = Vo - I(Ri)

Where Vt is terminal voltage, Vo is battery voltage, I is current and Ri is internal resistance. Is that correct? How would they do it?
Answered by bobpursley
current = 9.0/(internal resisitance + 50)
then
terminal voltage= 9.0-current*Ri
Answered by Mike
thank you very much
Answered by Mike
sorry here's another question ... they also ask what is the maximum amount of power you can get out of this battery?

You have P = IV. Max voltage is 9V, and max current is 2.7A. Then in that case that would be 9 * 2.7 = 24.3. Did I do that right?
Answered by bobpursley
No, not at all. Max power is delived when load resistance is equal to internal resistance.


Proof: Load power= Rl*Il^2=Rl(9/(Ri+Rl)^2

take the derivative:

9/(Ri+Rl)^2 - 18Rl/(Ri+Rl)^3
set that equal to zero, and solve for Rl

Rl=1/2 (Ri+Rl)
Rl=Ri for max power.
Answered by Damon
Yes, you short circuited the terminals. In all probability the battery will not last long but it sure will get hot.
Answered by Damon
However as far as useful power out of the battery instead of generating heat inside you had best look for where the external voltage times the current ins maximum.
There are no AI answers yet. The ability to request AI answers is coming soon!