0
$\begingroup$

Let's imagine there is a circuit that contains two identical cells connected wrong way round i.e. in reverse, and nothing else.

I know that normally one will therefore say that the emf of the two cells cancel and hence by Ohm's law no current flows- assuming that there is some resistance in the wires.

However, now let the situation be ideal such that the wires in the circuit have 0 resistance and the cells themselves also have 0 internal resistance. Therefore, there is no resistance anywhere in the circuit.

Now, were I to apply Ohm's law, I=V/R, then as R tends to 0, I will tend to infinity. Does this therefore mean that even though the p.d. is 0 in the circuit, the current is however infinite?

Is this the situation for a 0 resistance circuit containing only a cell and an inductor- with the cell's emf and the inductor's back emf?

Thank you for answering the question.

$\endgroup$
  • $\begingroup$ No. The total EMF is $0$, so there's no current, even at $R=0$. $\endgroup$ – Gert Nov 21 '17 at 15:40
  • $\begingroup$ @Gert, see my answer. $\endgroup$ – Alfred Centauri Nov 21 '17 at 16:35
0
$\begingroup$

Well, you may even consider your given system a non-circuit.

Imagine a battery with individual wire segments attached to either end so that wires stick out of each terminal. These wire segments aren't attached to anything but the battery's terminals. If you were to use a voltmeter to measure the voltage across the wires, you would see the voltage of the battery. In this situation, there is a potential difference from wire to wire although there is no circuit.

Your situation with the reversed batteries is similar. In an ideal condition, your batteries wouldn't even "think" it was in a circuit. The double battery setup just contributes to the potential difference across the wire segments and nothing else. You have simply made a larger battery.

If you are still not satisfied and require a mathematical solution, the limit of a function multiplied by 0 (however the divisor looks) will approach 0. In your case, the function isn't continuous, but a graph will show a limit at 0.

$\endgroup$
0
$\begingroup$

then as R tends to 0, I will tend to infinity

Why? If the ideal, identical cells are connected 'in reverse' (which, I assume, means that $+_1$ is connected to $+_2$ and $-_1$ is connected to $-_2$), there is no voltage across $R$ and so $I = 0/R$ which does not tend to infinity as $R \rightarrow 0$.

It turns out that, for $R=0$, the current $I$ is indeterminate, i.e., any value of $I$ is allowable. For example, insert an ideal constant current source between the batteries and find that the current $I$ is fixed by the current source and, further, that the voltage across the current source is zero for any value of $I$.

From this, conclude that the current $I$ is a free variable in the $R = 0$ case.

$\endgroup$
1
$\begingroup$

This may not be what you got caught up, but if you think of Ohm's law as V = IR, then 0 = 0 x 0 is a valid solution. Hence even if R=0, zero current is a good answer when V=0. (When R=0, you can't divide the above equation and get I=V/R).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.