college physics

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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?

• college physics -

.150*internalresistance=9-8.5

• college physics -

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?

• college physics -

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?

• college physics -

current = 9.0/(internal resisitance + 50)
then
terminal voltage= 9.0-current*Ri

• college physics -

thank you very much

• college physics -

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?

• college physics -

No, not at all. Max power is delived when load resistance is equal to internal resistance.

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.

• college physics -

Yes, you short circuited the terminals. In all probability the battery will not last long but it sure will get hot.

• college physics -

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.

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