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In my physics textbook there is a question about a circuit:

enter image description here

In question b) We find out the current in the circuit is 0.003A. This is before the voltmeter is added.

In question c) a voltmeter is added that affects the p.d across the first 1k resistor which causes its voltage to drop from 3.0V to 2.0V. In the mark scheme they say R(from B to A) = 666.7 ohms which is means they did 2/0.003. How is it known the current doesn't change at all, or am I missing something?

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  • $\begingroup$ Cant really get what you want an answer for. I would rather recommend that you first read about an ideal voltmeter. That should ideally solve your issue. $\endgroup$ – sbp Sep 12 '18 at 17:57
  • $\begingroup$ @sbp I am wondering how it is known the current across the resistor remains the same, even though the effective resistance of the circuit changed due to them adding the voltmeter $\endgroup$ – DevinJC Sep 12 '18 at 17:59
  • $\begingroup$ Shouldnt that ideally happen when you measure something? The measurement shouldnt ideally alter any observable. Now the question is how can a voltmeter really do this - should have a very high resistance! $\endgroup$ – sbp Sep 12 '18 at 18:01
  • $\begingroup$ @sbp Yes I understand that but the voltmeter in the question has a resistance of 2000Ohms (in the mark scheme) so why did the current in the resistor at BA not change $\endgroup$ – DevinJC Sep 12 '18 at 18:05
  • $\begingroup$ Could you just post the complete question? I mean the full text before the subpart begins. $\endgroup$ – sbp Sep 12 '18 at 18:06
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In the mark scheme they say R(from B to A) = 666.7 ohms which is means they did 2/0.003. How is it known the current doesn't change at all, or am I missing something?

The given solution is incorrect.

If we connect the voltmeter as shown and measure 2 V, then we know that $$ 6 \frac{R'}{R'+1000} = 2$$ where $R'$ is the parallel combination of the voltmeter and the 1 kohm resistor, from the voltage divider rule. This means $R'$ must be 500 ohms.

Therefore $$\frac{1000 R}{1000+R} = 500,$$ from which we can find $$R = 1000.$$

Sometimes the problem is just that the guy who wrote the textbook didn't pay much attention when writing the solutions.

How is it known the current doesn't change at all, or am I missing something?

A well-designed voltmeter should have a burden resistance of $10^6\ {\rm \Omega}$ or higher, so often it is reasonable to make this assumption. However, it should not have been made in this particular problem.

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  • $\begingroup$ My working did keep pointing towards 1000 but I assumed I was wrong lol. Thanks for the clarification :) $\endgroup$ – DevinJC Sep 12 '18 at 18:30

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