# Why is there a voltage drop across things when no current is flowing?

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.

• What is the resistance of the open switch? If you claim you can use Ohm's law, you must be able to produce not only the $I$, but also the $R$ for it. Commented Nov 8, 2016 at 14:35
• Well I guess you're leading me to the point that the resistance is infinite, and so is it just the fact that Ohm's law is not applicable here? Commented Nov 8, 2016 at 14:41
• Think along the lines of Kirchoffs loop rule Commented Nov 8, 2016 at 14:45
• voltage has nothing to do with the current...voltage causes current to flow where that is possible, it does not mean that it needs current to exist. Voltage is just the potential difference betwen two points in space. Commented Nov 8, 2016 at 14:56
• Ah - but doesn't your measuring instrument draw current when you connect it to the potential - even though it's of high impedance it indeed draws electrons to register! Commented Nov 8, 2016 at 15:38

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.

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.

Using the water analogy of circuits, this is like asking how two sections of connected pipe could be at different heights when there is no water flowing between them.

This is just a conceptual mistake. The Ohm's law states that the voltage drop developed across a conductor is proportional to the current flowing through the conductor, the proportionality being a property of the conductor itself- it's resistance. This is what we write mathematically as

$$V=IR$$

Now, when you connect an open switch and a resistor to the battery, the open switch allows infinite resistance to be there in the circuit. Hence the current flowing through the circuit will be

$$I=\frac{V}{R}=\frac{V}{\infty}=0$$

This means no current flows through the conductor. But the applied voltage has to be there somewhere. It has to be so. So the entire voltage appears across the resistor.

Now, what happens if you apply voltage across a conductor with zero resistance? Then the voltage will be

$$V=IR=I_\textrm{max}(0)=0$$

where $I_\textrm{max}$ is the maximum current that flows through the circuit.

These arguments can be visualized so easily as follows. Imagine the circuit. The switch is open. This means, the charges could not flow through the circuit and complete the circuit. This is due to the gap in between the switch. So the applied energy is not converted to current even though the resistance is still there. Hence voltage will be there, where you applied it.

Now, in the second case, there is no resistance. This means the supplied energy (the voltage) is completely utilized by the charges to flow through the circuit. There is nothing left so as to be detected by any external circuit (a voltmeter probably). Hence one cannot see any voltage difference between the conductor. You can see energy somewhere if it is there. But if it's being utilized, then you will not see it there, but its there in some other form- here it is the kinetic energy of the charges.

All we invoked here is the energy conservation (not precisely...).

Just for ease let's assume that the wire, the resistor and the switch are all ohmic conductors with resistances $R_{\rm wire}, R_{\rm switch}$ and $R_{\rm resistor}$.

For this circuit $12 = I(R_{\rm wire}+R_{\rm switch}+R_{\rm resistor})$ and so the voltage across the switch is

$\dfrac {12\;R_{\rm switch}}{R_{\rm wire}+R_{\rm switch}+R_{\rm resistor}}$.

When the switch is open $R_{\rm switch} \gg R_{\rm wire}, R_{\rm resistor}$

So the voltage across the switch $\approx 12 \rm V$ and the current $\approx\dfrac{12}{R_{\rm switch}} \approx 0 \rm A$ if $R_{\rm switch} \gg 12$.

If the switch is closed then $R_{\rm resistor} \gg R_{\rm wire}, R_{\rm switch}$

So the voltage across the switch $\approx \dfrac {12\;R_{\rm switch}}{R_{\rm resistor}}\approx 0 \rm V$ and the current $\approx \dfrac {12}{R_{\rm resistor}}$.

Because there is no current the drop on the resistance is zero, that means that the voltage difference at the ends of the switch is the same as the ends of the battery.

voltage change /potential difference actually means the potential energy that can be converted to other form.now while in conversion, applying OHM'S LAW , V=IR here R(resistance) actually consumes the potential energy when charges flow through a conductor. so when R is absent in an IDEAL CLOSED SWITCH, THEN THE WHOLE POTENTIAL ENERGY ,IS CONSUMED THUS THERE IS NO POTENTIAL CHANGE