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Can someone please solve this for me, telling me what's the right option and why? I will translate for you the question:

"What is the reading shown by the voltimeter inserted in the circuit of figure 2".

R stands for resistence and V for volts.

It wasn't me that selected the option B and I am not sure if it really is the right option.

Question I want to learn how to solve

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  • \$\begingroup\$ Use the fact that a Voltmeter has a large resistance. \$\endgroup\$ – ijuneja Jun 9 '18 at 15:59
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    \$\begingroup\$ You were correct to be suspicious that B is not the right option. It isn't. You're also correct that (for an ideal voltmeter) there'd be no current flowing. So now ask yourself: if there's no current flowing, what would the voltage drop across R1 and R2 be? And after that the voltmeter must measure what remains... \$\endgroup\$ – brhans Jun 9 '18 at 16:12
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    \$\begingroup\$ 9V, because if there is no current flowing(resistence = +infinite), there is no voltage drop, so it reads the same as the source? \$\endgroup\$ – Tiago Cascais Jun 9 '18 at 16:18
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    \$\begingroup\$ Sooo about the current flowing. I am seeing that, we're just talking about one terminal to another and assuming that is an ideal voltimeter, right? \$\endgroup\$ – Tiago Cascais Jun 9 '18 at 16:18
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    \$\begingroup\$ An ideal meter is open, as an ideal meter will introduce no error in the measurement. Assuming ''very high compared to" is a next step into making it more realistic, but may (unnecessarily) confuse a student. \$\endgroup\$ – Joren Vaes Jun 9 '18 at 17:09
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In order to measure voltage, you place the meter in parallel with the element that you are trying to measure the voltage across so the meter must have very high, or in an ideal case infinite, resistance to avoid drawing current and messing up your measurement. So for simplicity let's assume we have an ideal voltmeter with infinite resistance. Now, you can analyze the circuit using Kirchhoff's voltage law (KVL).

schematic

simulate this circuit – Schematic created using CircuitLab

Starting from the battery.

\$9V - I_{Loop}\times R_{1} - V_{Voltmeter} - I_{Loop}\times R_{2} = 0\$

Since the voltmeter has infinite series resistance, we can model it as an open circuit thus \$I_{Loop} = 0\$.

Since \$I_{Loop} = 0\$

\$9V - 0\times R_{1} - V_{Voltmeter} - 0\times R_{2} = 0\$

Or

\$V_{Voltmeter} = 9V\$

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    \$\begingroup\$ Nicely stated.+1 \$\endgroup\$ – Blair Fonville Jun 9 '18 at 18:58
  • \$\begingroup\$ Thank you all for dispending your precious time with me :) :) \$\endgroup\$ – Tiago Cascais Jun 9 '18 at 21:32
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The circuit looks like this when you consider that the voltmeter has a resistance itself:

schematic

simulate this circuit – Schematic created using CircuitLab

Here I assumed the voltmeter has a resistance of 100K Ohms, which if it is 10000 times the R values works out to R=100 Ohms. A moving coil voltmeter might have an R value of 100K Ohms, whereas a modern FET based digital voltmeter more likely in the 10M Ohms range.

The voltage across the battery is fixed at 9 V if we ignore internal resistance and we have a simple voltage divider across the battery.
The voltage across Node A-B is

NodeA-B = 9V * 100000/100300

        = 8.97 V

The answer is very close to 9 V in this instance with these values, and if you consider the voltmeter as 'ideal' ( R tends to infinity) the NodeA-B voltage will tend ever closer to 9V.

You should read up on the voltage divider rule to understand the formulae involved.

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    \$\begingroup\$ "if you consider the voltmeter as 'ideal' (R tends to infinity) the NodeA-B voltage will tend ever closer to 9V." Ideal voltmeter does not "tend to infinity", it IS an infinity. So, you can put 100M resistors there and still get 9V. I actually agree with other comments that in your attempt to be close to reality you've missed the mathematical aspect of the problem. \$\endgroup\$ – Maple Jun 9 '18 at 17:11
  • \$\begingroup\$ Thank you all for dispending your precious time with me :) \$\endgroup\$ – Tiago Cascais Jun 9 '18 at 21:32

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