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?
9 answers
.150*internalresistance=9-8.5
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?
Now they ask, in the second part .... What would be the terminal voltage for a 50-W load?
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?
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?
current = 9.0/(internal resisitance + 50)
then
terminal voltage= 9.0-current*Ri
then
terminal voltage= 9.0-current*Ri
thank you very much
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?
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?
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.
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.
Yes, you short circuited the terminals. In all probability the battery will not last long but it sure will get hot.
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.
3 answers