Asked by H

11. If you have a circuit set up where you have have a 9 V battery and one resistor of 10 Ohms. You place a voltmeter across the two ends of the battery and it reads 8.333 V. What is the internal resistance of the battery?

12. A 12 V battery is set up in an unknown configuration. The internal resistance of the battery is 4 Ohms. If this circuit must have a minimum of two resistors, what resistance must these resistors have and what configuration must they be in in order for a terminal voltage to read 11.478 V? Show all work
Answered by henry2,
11. Assuming the no-load voltage is 9 volts:
I = V/R = 8.333/10 = 0.833A.
r = (E-Vl)/I = (9-8.333)/0.833 =

12. Given: E = 12V., r = 4 ohms, Vl = 11.478V.
I = (E-Vl)/r = (12-11.478)/4 = 0.131A.
Req = V/I = 11.478/0.131 = 88 ohms. = R1 and R2 in-parallel.
R1 = R2 = 2*88 = 176 ohms each.
Answered by TGS
How did you know its in a parallel? I believe that you are right btw.
Answered by henry2,
Series connection: R1 = R2 = 88/2 = 44 ohms.
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