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Consider a simple circuit with a battery of constant potential difference V and a resistor of resistance R. So if we close the circuit then current flows through it and there is a voltage drop across it. As there is only one resistor the total voltage drops across it. Which means if we measure the potential difference using a voltmeter then it shows 5V(considering the conducting wire has zero resistance). So on one end the potential is 5V and on the other end it is 0V. Now my question is how the current flows further in the circuit from the 0V end as there is no potential difference between this end of resistor and the ground. I know I have missed something basic, so please help me out.

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In your idealised scenario you are assuming that the wires have zero resistance, so no potential difference is required to drive current through the wires.

In practice the wires will have a small non-zero resistance, so there will be a small voltage drop across each wire. The potential at one end of the resistor will be slightly below $5$V and the potential at the other end of resistor will be slightly above $0$V.

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  • $\begingroup$ So if the potential at one end is slightly below 5V then why the whole potential didn't drop across the resistor instead there is potential slightly above zero. $\endgroup$ – lfp Jan 30 '20 at 15:09
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    $\begingroup$ @lfp When you model wires as having a small non-zero resistance, you end up modeling two small resistors (wire resistances) in series with the resistor you are testing. Solving that equation provides the drop across each wire. Of course, when talking about practical applications, you'll have to make sure other effects don't play a part like the resistance of your power source. In the setup you describe, that resistance wont matter, but if you start adjusting the experiment to measure these tiny voltage drops, it may start to matter. $\endgroup$ – Cort Ammon Jan 30 '20 at 15:13
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When the charge returns to the battery having zero electrical potential, it gains 5 volts electrical potential going from the negative battery terminal to the positive due to the conversation of chemical potential energy to electrical potential energy inside the battery.

Hope this helps.

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  • $\begingroup$ Same as connecting the resistor directly to the negative terminal with no wire. $\endgroup$ – user45664 Jan 30 '20 at 16:55

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