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enter image description here

My questions will be based on the above figure.

  • The Emf Source V1 and R1 Together form a cell(R1 is internal resistance)
  • R2 Represents the entire load on the source

VoltMeter

My teacher told us that Emf was the terminal potential between the electrodes of the source when the ciruit is open and the current drawn is zero.

Earlier in the class he told us that a volt meter is considered to have $\infty$ Resistance so as to draw nearly zero amperes from the loop.

Next He told us that the current through the Resistor R1 is zero as the circuit is open do the reading on the volt meter will only be the potential of the Source V1 i.e. Emf

My question is

  • If the circuit is open and the current through R1 is zero and also the voltmeter is drawing zero current then how could the voltmeter possibly show any reading at all?
  • How can one measure the voltage if the circuit is open .
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This is Catch 22 type of question. Of course resistance of the voltmeter is never infinitive and of course some current does flow through the voltmeter. The idea is that resistivity of the voltmeter is much larger than other resistivities, i.e.

$$R_V \gg R_1, R_V \gg R_2.$$

If switch SW1 is open, then current through voltmeter equals $\frac{V_1}{R_V+R_1}$ and voltage on the voltmeter thus equals

$$V = \frac{R_V}{R_V+R_1} V_1 \approx V_1.$$

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  • $\begingroup$ and what about when the switch is closed $\endgroup$
    – The-Ever-Kid
    May 21, 2012 at 10:47
  • $\begingroup$ $R_\text{total} = R_1 + \frac{R_V R_2}{R_V + R_2} \approx R_1 + R_2$, then $I=V_1/R_\text{total}$ and $V = I R_1 = \frac{R_1}{R_1+R_2} V_1$ $\endgroup$
    – Pygmalion
    May 21, 2012 at 10:53

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