Why would it be wrong to calculate the voltmeter reading in the following diagram as one plus 2 ohms is 2/3 ohms as they are in parallel, hence the p.d. reading on the voltmeter must be the current times the resistance which is 1.5 V. Why does this not work?

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  • $\begingroup$ How do you decide if two resistors are in parallel? $\endgroup$
    – Puk
    Commented May 4, 2022 at 8:09

2 Answers 2


Instead of presenting a solution, I'd like to point out where you're going the wrong way.

You're talking about "the" resistance and "the" current, as if there were only one resistance and one current in the whole circuit. You need to be much more explicit in expression (and probably also in thinking) when talking about circuits.

The circuit has:

  • 4 resistors, each having its own resistance value, a voltmeter, which might or might not have a resistance value. By simplifying the circuit, you might introduce additional, computed resistances.
  • 6 branches (battery branch, 4 resitor branches, voltmeter branch), each with one current value flowing throught the branch, and one voltage across the branch.
  • 4 nodes where the branches meet (left-hand and right-hand end of the bridge plus the two nodes in between, where the voltmeter is connected.

If you want to be clearly understood, without the risk of mis-interpretation, it's a good idea to give individual names to all of these circuit elements.

With "one plus 2 ohms" you probably mean the two resistors to the left (but that's already guesswork...).

You write "as they are in parallel". That isn't true. To be parallel, they need to be both at the same network nodes, at both ends. To the left, they share a node, but to the right, they are connected to different ends of the voltmeter, so they are not directly connected there.

The "parallel resistance rule" isn't only a formula to do some calculations, but also defines pre-conditions and states what to do with the result.

If you have two resistors that are directly connected to one another at both ends, then you can effectively replace them with a single resistor whose value is calculated as $R_p = \frac{R_1 \cdot R_2}{R_1 + R_2}$

So, you can understand it as a redrawing rule, that allows you to replace a circuit with a simpler one, having the same effect to the outside.

You tried to apply this rule although its preconditions weren't fulfilled.

If the voltmeter were a direct connection (behaving like an ideal wire), then you could treat the resistors as being parallel. But then there would be no voltage difference between the two ends of the voltmeter, and it would always show zero, rendering it useless. And that's why an ideal voltmeter does not behave like a wire but like a resistor with infinite resistance, effectively removing its "connection" from the circuit.

You write "...must be the current times the resistance which is 1.5 V". Which current? Again, you apply a rule (Ohm's law) without ckecking whether its preconditions are met.

The voltage U across a resistor is the current flowing through this resistor times its resistance value.

The formula is true only if you take voltage, current and resistance at one single resistor, not if you mix a current from here with a voltage from there and the resistance from still another place.

If you want to directly use Ohm's law to compute the voltage across the voltmeter, you need

  • the resistance value of the voltmeter (unknown, typically assumed to be infinite for an ideal voltmeter),
  • the current through the voltmeter (hard to compute without a given voltmeter resistance, typically assumed to be zero for an ideal voltmeter),
  • an assertion that the voltmeter behaves like a resistor (not necessarily true).

At best, you end up with $U = \infty \cdot 0$, which can be any number you like.

  • $\begingroup$ Hey thanks so much!! So you mean that the voltmeter reading can be any number? The answer is 1 V $\endgroup$
    – Blue Green
    Commented May 4, 2022 at 12:55
  • $\begingroup$ No, of course the 1V reading can be solidly computed. I just wanted to stress that it's more complex than trying to apply a random formula (Ohm's law) to some random values. The "art" of solving circuit-related questions is to wisely select which rule to apply where. The important questions are: where can I apply which law, which numbers do I have to insert into the formula, and what does the computed result tell me? $\endgroup$ Commented May 4, 2022 at 14:23

Yes, you can imagine the upper and lower branch to be in parallel. The assumption here is that the voltmeter is ideal, that is only a negligible current flows through it.

But draw the upper and lower branch separately first as two loops. As we established that no current flows between the two branches. Then what is the potential difference as you go along these two individual loops at each point?


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