electric-circuits
If you have an electric circuit with a 12V battery in series with an open switch and a resistor, the voltage drop across the open switch is 12V. But this doesn't quite make sense to me. If there is no current, why does Ohm's Law not apply giving me a voltage drop of V = IR = 0 as there is no current?
I guess more generally I'm confused as to why things with zero current going through them have a voltage drop at all as V=IR.
The answer is simple, the resistor doesn't know what voltage drop to provide, and to some extent it doesn't "care" either (unless it's so high that the current fries it but that's another issue).
The "voltage drop" is only governed by the potential difference created by the generator (battery or other). If the circuit is open, the electrons have no path so there is no flow. As soon as the circuit is closed, the electrons have a path and start to flow across the conductor (a resistor is also a conductor).
The only thing the resistor will govern, is how strong the flow of electron will be from A to B (from one potential to another, but remember this potential was created by the generator, the resistor doesn't need to know it).
I often do the analogy between electricity and hydraulic flow. Consider the difference of potential (the voltage) created by the generator as a height, and the resistor as a slope. The flow of water will be different in the same way as the electrons in a circuit.
Consider the 3 scenario in the following image:
A:
No resistor to close the circuit => no current flow possible.B:
Low resistor. The electrons (or the water) fall from the high potential to the low one (rather rapidly).C: The electrons (or the water) fall from high to low potential, still from the same height (your Voltage drop is identical). Except this time the higher resistance (flatter slope) makes it harder to reach the low potential (=> lower current).Now imagine the high potential is 24V, or even 10,000, the mechanics are still the same, the electrons will flow from one potential to another as soon as they have a path, regardless of the resistor value. The only difference made by the resistor is how strong they will flow (how strong the current will be).
note: The analogies with the water flow are easily arguable and quickly reach their limits with complex circuits, this is not the exercise here. They still are a great way to explain electric current in the simple cases
The circuit is drawing infinite amount of current from the battery. Have I correctly understood this ?
Yes.
There will be a potential difference across the resistor in parallel to capacitor and that potential difference will be resposnsible for charging it
The potential across the capacitor can't change instantaneously.
Therefore in the time immediately after the switch closes, the voltage across the resistor (the one in parallel with the capacitor) is zero. From Ohm's law, then, there is no current through this resistor in that instant.
To find the current that is charging the capacitor (in the instant immediately after closing the switch), you can use KCL at the node where the capacitor and the two resistors are all connected.
Alternately, you can replace the voltage source and the two resistors with a Thevenin equivalent circuit, and again find the charging current as time progresses after the switch is closed.
Best Answer
Ohm's law applies to ohmic devices; if the voltage across a device is proportional to the current through, the device is ohmic otherwise it isn't.
Ohm's law is not a universal law. For example, Ohm's law does not apply to capacitors, inductors, diodes, transistors, vacuum tubes, etc. etc.
An open (ideal) switch is not an ohmic device since the current through $(0\mathrm{A})$ is not proportional to the voltage across. However, one can think of an open switch as the limit as $R \rightarrow \infty$ of a resistor.
A closed (ideal) switch is not an ohmic device since the voltage across $(0\mathrm{V})$ is not proportional to the current through. However, one can think of a closed switch as the limit as $R \rightarrow 0$ of a resistor.