Recently I encountered a circuit problem. Imagine, you have an ordinary circuit with a resistor and a battery($60\,\mathrm{V}$ emf and negligible internal resistance denoted by r) with a switch($S_1$) connected in parallel (Call the end points of this wire AB).

The picture of the circuit is in the diagram below:-


{To restate, the resistor and the switch $S_1$ are in parallel with a battery of $60\,\mathrm{V}$ emf.}

Once you have drawn this, the question is as follows: If $S_1$ is open then what will be the potential difference across the endpoints of the switch.

The Answer is $60\,\mathrm{V}$ but how is it possible? Since no current will flow through the second branch of the circuit, by Ohm's law $V=IR$, if $I=0$, then shouldn't this imply that $V=0$?

Please give a logical explanation.


In the following situation

Circuit Diagram 1

you have the voltage source ensuring a potential difference V = 60 Volts between its terminals. The source's upper terminal is connected to the switch's upper terminal, so they have the same electric potential.

The switch's lower terminal is connected to the resistor's upper terminal, so they also have the same electrical potential. As you correctly stated, there is no current flowing through the resistor, so by Ohm's Law the voltage difference across the resistor's terminals is 0. Therefore, the resistor's upper terminal has the same electric potential as the resistor's lower terminal.

However, the resistor's lower terminal is also the source's lower terminal, which has a potential difference of 60 Volts with the switch's upper terminal.

Therefore, the potential difference across the switch is 60 Volts, even though there is no current flowing through it.

An open switch can be modelled as a resistor with infinite resistance, so if you apply Ohm's Law directly to it, you can have a potential difference even though the current flowing through it is zero.

In the following situation

Circuit Diagram 2

you have that the ideal voltage source always assures the 60 Volts potential difference between its terminals, regardless if the switch is open or closed. Therefore, there will always be a current $I = \frac{V}{R}$ flowing through the resistor, and if the switch is closed, there will be an infinite current flowing through it (assuming the switch's resistance as zero).

In practice, what would happen is that the current flowing through the switch would be very large, and the wire would melt; I've seen it happen a few times when my students accidentally short-circuit the source in my Circuits Lab class.


The problem here lies with the assumption that the resistance of the battery, wire and switch are negligible. There are two solutions that I can think of:

1) Because the two branches are connected in parallel, the branch with the resistor will still have $I_R=\frac RV$ current through it while the switch $S_1$ will have "infinite" current going through it as the voltage is still set to $V$ (that is because the switch's resistance is zero, which is unrealistic and therefore poses a problem).

2) The battery has an internal resistance of $r$, therefore the current through the wire when the switch is open will be $I_{AB}=\frac rV$, but all of the $60V$ will be dropped upon the battery itself, and the voltage drop on the wire will be 0.

In general, neglecting resistances sometimes give unrealistic results.


The solution is very simple.

[EDIT: My diagram is almost similar to yours, just without the internal resistance r. I have talked about r in the last few lines. Anyway my diagram totally logical and valid for your problem.]

See, your circuit sums up as follows:

enter image description here Now since the switch is open, you can consider that the lower branch of the circuit has been detached and taken away by someone.

(This consideration comes from the viewpoint that an open branch of an electrical circuit has an infinite resistance as it has to cross the intervening medium "air" between the open ends of the switch and air is a very very good insulator of electricity.)

But the circuit is still complete as the other branch with the resistor is intact . So current through the circuit now passes through the resistor as its resistance is less compared to (the infinite resistance in) the other branch.."open switch, I mean."

The circuit becomes like this: enter image description here

Hence the entire potential difference builds up on the resistor...that is , $60\,\mathrm{V}$ builds up across AB.

However since your diagram shows that the battery has some internal resistance, then obviously the answer won't be $60\,\mathrm{V}$. It would be less. The value of $V$ across AB will be $(60-Ir)\,\mathrm{V}$ .


Since, the switch is connected to both the terminals of resistance by a wire and the wire must have negligible resistance so the both terminals of switch and resistance are at same potential or better to say that these are equipotential surfaces. . ------.1. .2.----- . | | . | | . 5------3///\4----6 . |. | . |. | . ---battery-///---

Lets consider this figure In this figure 1,3&5 and 2,4&6 are at separately are at same potential so according to the definition of equipotential surfaces we can merge or separate the points which are at same potential so the figure can be transformed to

. 1,3&5------///------2,4&6 . |. | . |. | . ---battery--///-- Therefore the potential difference between the terminals of switch will be same as that of the resistor.

  • $\begingroup$ This could use significant editing to clarify. As it is it is impossible to follow what you mean. $\endgroup$
    – Jon Custer
    Jun 27 '16 at 17:06

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