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enter image description here

This is the basic simple question. When I connect this circuit there is voltage seen in VM2. According to me, no current flows through R5, therefore there should be no voltage across the resistor, and resistor should act as open. Therefore VM2 should be zero.

Can anyone help me why it is like this?

Well I appreciate your answers. If I use simple resistor divider with the diode whose forward voltage drop is 0.6V, I am reading voltage in voltmeter as 3.88V. Is this because of any other phenomenon, can somebody please explain.enter image description here

It should have read 2.5V.

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  • \$\begingroup\$ Short answer: a resistor with no voltage across can be treated as a short circuit, not an open circuit. \$\endgroup\$
    – Joe Hass
    Dec 16 '13 at 12:59
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Therefore there should be no voltage across the resistor, and resistor should act as open. Therefore VM2 should be zero.

An open circuit can have any voltage across with zero current through. However, a resistor can only have zero volts across with zero current through so your reasoning is flawed.

But, this is actually a very common reasoning error among those learning electrical circuits. To develop correct intuition, you must check it against the most fundamental circuit laws.

First, there's KVL. We can write a KVL equation around the rightmost loop:

$$V_{R1} = V_{AM1} + V_{R5} + V_{M2}$$

This must hold. If your intuition violates KVL, your intuition is leading you astray.

Second, we have Ohm's Law:

$$V_{R5} = 0A \cdot 100\Omega = 0V$$

Also, \$V_{AM1} = 0V\$ since it is an ideal ammeter.

Thus, regardless of your intuition, it must be the case that

$$V_{R1} = V_{M2}$$

The voltmeter reads the voltage across the resistor R1.

To help develop your intuition, think about moving the "top" voltmeter lead to the leftmost terminal of R5. I think it is clear that the voltmeter will read the voltage across R1.

Now, since there is zero volts across R5, the voltage reading on either side of R5 must be the same. In other words, moving the voltmeter lead back to the rightmost terminal of R5 should not change the voltmeter reading at all.

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  • \$\begingroup\$ Nice answer, and congrats on your 10k. \$\endgroup\$
    – PeterJ
    Dec 16 '13 at 12:40
  • \$\begingroup\$ Thanks @AlfredCentauri for your answer. Can you please tell me the answer for second part i.e diode circuit. \$\endgroup\$ Dec 18 '13 at 4:21
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R5 is a resistor and always behaves as a resistor, not an open circuit. As a resistor it obeys Ohm's law, V=IR. If I (the current through it) is zero, then V (the voltage drop across it) is zero. As no voltage is dropped across R5, the voltage measured by VM2 will be the same as the voltage across R1.

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We make the usual simplifying assumptions here a current meter has zero resistance and a volt meter takes no current.

R5 therefore has no current through it so there is no volt drop across it.

The voltage as measured by VM2 will therefore be the same as if you were to remove AM1 and R5 and measure directly across R1. This is a simple potential divider the voltage is thus:

$$V2 \cdot \frac{R1}{R1+R2+R3} = 5 \cdot \frac{100}{300} \approx 1.667 \text{ volts} $$

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According to me, no current flows through R5, therefore there should be no voltage across the resistor (R5)

That is correct, the voltage drop across a resistor increases as the current that flows through it increases, when the current through it is 0 then the voltage drop across it will also be 0.

and resistor should act as open. Therefore VM2 should be zero.

That is the wrong part, when you have a resistor and you connect a voltage in one side of it then if there is no voltage drop across it (like in your case because no current flows) there is no alternative but to have the same voltage in the other side of the resistor too, the voltage can't disappear.

This can be explained mathematically by Ohms law

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