Why does a potentiometer measure emf better than a voltmeter, and what actually increases its sensitivity?

7 answers & replies

• [+][–]Arjun Nair12thSample· 1mo ago

  1. At the balance point, the potential difference across the length $l$ of the wire exactly equals the emf of the cell, so the two push equally against each other and the galvanometer carries zero current. That means the cell neither delivers nor takes current, so we are measuring its actual emf. A voltmeter always has finite resistance, draws a small current, and the terminal voltage it reads is $E - Ir$, a little less than the emf.

  2. Along the wire, the potential drops from the driver's full value to near zero. For the balance point to exist somewhere along the wire the drop across the whole wire must be at least $E$. If the driver emf is smaller there is no point where the drop equals $E$ and the galvanometer never reads zero.

  Report
• [+][–]Rahul DubeyDropperSample· 1mo ago

  3. Internal resistance: $l_1$ is the balance length with the key open (cell on no load, so you are measuring emf). $l_2$ is the balance length with the key closed (cell sends current through the known resistance $R$, so you are measuring terminal voltage $V$). Since $E = kl_1$ and $V = kl_2$, and $V = E - Ir$ with $I = V/R$, you get $r = R\left(\dfrac{l_1 - l_2}{l_2}\right)$.

  Memory hook: $l_1$ comes first and is bigger (open circuit gives the larger voltage), $l_2$ is the smaller one.

  Report

  • [+][–]Zoya Khan12thSample· 1mo agoAuthor

    The "l₁ is bigger" hook helped. I couldn't remember the order before. Thank you for the patient answers everyone.

    Report
• [+][–]Harsh Gupta12thSample· 1mo ago

  Sensitivity: increase the current in the main circuit by reducing the rheostat resistance, so the wire has a bigger voltage drop and the balance point is clearer.

  Report

  • [+][–]Kavya NaikDropperSample· 1mo ago

    Harsh, I think that's the opposite! Sensitivity improves when the potential gradient $k$ is smaller, because a small change in jockey position then corresponds to a very small change in potential and you can balance more precisely. So reduce the current in the main circuit (add resistance via the rheostat), or use a longer wire. The trade-off is that the driver's drop across the wire must still be larger than the emf you want to measure.

    Report

    • [+][–]Harsh Gupta12thSample· 1mo ago

      Oops. I got it backwards, I was thinking "more current more signal". Kavya didi is right: smaller gradient, finer reading. Moving to a career in standup where nobody checks the physics.

      Report
• [+][–]Vivek Mishra12thSample· 1mo ago

  Exam note. Two advantages that come up in statement questions: it is a null method, so no current is drawn from the cell being measured, and its sensitivity can be raised by lowering the potential gradient (longer wire). The main disadvantage is that it needs a long wire and a steady driver cell, so it is not handy for quick readings.

  Report