legrandfromage123 Posted September 30, 2012 Report Share Posted September 30, 2012 A cell is connected in series with a 2.0 Ω resistor and a switch. The voltmeter is connected across the cell and reads 12 V when the switch is open and 8.0 V when the switch is closed.The answer is supposed to be 1 Ω, but I can't seem to get there! I've tried using the formula V=emf-Ir but i seem to be doing something wrong.Any help with this question would be appreciated! Reply Link to post Share on other sites More sharing options...
SerUmbras Posted September 30, 2012 Report Share Posted September 30, 2012 Exactly what is the question? I mean, what are you trying to find? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 1, 2012 Report Share Posted October 1, 2012 A cell is connected in series with a 2.0 Ω resistor and a switch. The voltmeter is connected across the cell and reads 12 V when the switch is open and 8.0 V when the switch is closed.The answer is supposed to be 1 Ω, but I can't seem to get there! I've tried using the formula V=emf-Ir but i seem to be doing something wrong.Any help with this question would be appreciated! When the switch is closed the current in the circuit is 8/2 = 4A and the voltage across the cell falls from 12V to 8V. Hence the internal resistance of the cell is (12 - 8)/4 = 1 ohm. Reply Link to post Share on other sites More sharing options...
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