A 1.40-OLHMS resistor is connected across a 9.00-V battery. The voltage between the terminals of the battery is observed to be only 8.50 V. Find the internal resistance of the battery.
2007-04-01 13:42:40 · 3 answers · asked by neverconsiderlifeonascale 1 in Science & Mathematics ➔ Physics
8.5 volts / 1.40 ohms = 6.07 amps
(9 volts - 8.5 volts) / 6.07 amps = .082 ohms
2007-04-01 13:50:43 · answer #1 · answered by Thomas C 6 · 1⤊ 1⤋
Two things that might help you understand what's happening: first, the current through all devices connected in series is the same. Second is that the voltage drop is proportional across any resistance in the circuit. The second is merely a consequence of Ohm's law. So if an EMF of 9V is put across two resistances and one has 8.5V across it, the other has 0.5V across it. The other answers (0.082 ohms) are correct, but play around with it a bit. It's the only way you learn.
2007-04-01 14:15:19 · answer #2 · answered by Anonymous · 1⤊ 0⤋
Ok :
You need to remember this :
E - I*r = V
r = internal resistance
V = voltage between the terminals
I = current through the resistor
I = 8.5 / 1.4 amp
then :
9 - 8.5/1.4*r = 8.5
9 - 6.07r = 8.5
r = 0.08 Ohms
Hope that helps
2007-04-01 13:50:15 · answer #3 · answered by anakin_louix 6 · 1⤊ 0⤋