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enter image description here this is an a level physics question, can someone here please explain why voltmeter reads zero and how changing resistance of variable resistor effects voltmeter reading. The main concept i want to really know how connecting batteries and resistors together can help to give different ranges of voltages across resistors in circuit, as this thing is quite different from the basic concept where batteries were connected together(in series or parallel). This question,however, requires different sort of understanding.

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closed as off-topic by ahemmetter, M. Enns, Bob D, Pieter, Dale May 16 at 21:49

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  • $\begingroup$ yes you are right but thats a different case, in that case no current would flow but, this case is different, i hope you can help. $\endgroup$ – Syed Abdullah May 16 at 19:42
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can someone here please explain why voltmeter reads zero and how changing resistance of variable resistor effects voltmeter reading

The voltmeter will read zero when the voltage across the 4 ohm resistor is 6V (top-most terminal positive). The voltage across the battery and 4 ohm resistor is then the sum 6V and -6V which equals zero volts.

Now, this requires that the current $I$ (down) through the 4 ohm resistor equal 1.5A (Ohm's law), and the current $I$ depends on the total resistance in the circuit:

$$I = \frac{6V + 6V}{4\Omega + 1.5\Omega + R}$$

Setting $R = 0$, the current is $I = \frac{12V}{5.5\Omega} \approx 2.2A$. Increasing $R$ decreases $I$, and so there is a value for $R > 0$ that gives a current of 1.5A

This is a perfectly general approach that applies even if the two batteries had different voltages. As another answer points out, due to the symmetry of this circuit, one can find the value of $R$ that gives a zero volt reading by inspection.

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The question does not say the voltmeter reads zero irrespective, but that at a certain resistance of R it will read zero.

The answer is most easily seen from symmetry. At what resistance of R will the circuit on the left hand side of the voltmeter be equivalent to that of the right hand side of the circuit?

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From the diagram we can understand that the potential is the same throughout $AB$ so we can easily deduce that the terminals of the voltmeter be at $+6~V$(taking conventional current direction) which means out of $12~V $ of potential difference created by the batteries, $6~V$ are used up between the points $B$ and $F$ which gives us the $1^{st}$ equation, $$6=I(x+1.5)$$ and that implies the remaining $6~V$ will be consumed by the $4\Omega$ resistor which gives us the $2^{nd}$ equation, $$6=4I$$ Solving them gives us $$x=2.5\Omega ~~\text{ and }~~I=1.5\text A$$ Here is a picture for your reference.

enter image description here

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