Question

Figure shows a simple potentiometer circuit for measuring a small emf produced by a thermocouple.
The meter wire PQ has a resistance of 5 ω, and the driver cell has an emf of 2.00 V. If a balance point is obtained 0.600 m along PQ when measuring an emf of 6.00 mV what is the value of resistance R?

A. $95\omega $
B. $995\omega $
C. $195\omega $
D. $1995\omega $

Answer 1

Hint: By using potentiometer we can measure the internal resistance of the cell or unknown emf of the cell. There will be voltage divided uniformly across the potentiometer wire and by balancing the cell at the length where the emf of the cell will be equal to potential difference across that portion of wire, we can find out unknown emf.
Formula used:
${R_s} = {R_1} + {R_2} + {R_3} + .........$
$V = i{R_{eff}}$

Complete answer:
Rate of flow of charge is known as current. A material which allows current to pass through it is known as a conductor. No conductor will be perfect. It will have some resistance. The property to hinder the flow of current is called resistance and a device which does that is known as a resistor.
If the same current is passing through all resistors then we tell those are connected in series. If potential difference is the same for all resistors then those resistors are told to be in parallel.
Same current will be passing through the external resistor and potentiometer wire as they will be in series. Let the resistance of the external resistor be R and the resistance of the potentiometer wire is given as $5\omega $ ohms. So the effective resistance will be
${R_s} = {R_1} + {R_2} + {R_3} + .........$
$\therefore {R_s} = R + 5\omega $
So the current through potentiometer circuit will be
$V = i{R_{eff}}$
$\eqalign{
  & \Rightarrow i = \dfrac{V}{{{R_s}}} \cr
  & \therefore i = \dfrac{2}{{R + 5\omega }} \cr} $
Since resistance will be divided uniformly throughout the potentiometer wire and the current passing through the potentiometer wire is constant, the potential difference also will be divided in proportion to resistance. For clear understanding one can refer the below diagram

Let ‘i’ amount of current passing through wire. $5\omega $ ohm resistance for 100cm means that every 10cm has $0.1 \times 5\omega $ohm resistance. So 60 cm has $0.6 \times 5\omega $ohm resistance. So potential difference across the 60cm wire will be
$V = i{R_{eff}}$
$\eqalign{
  & \Rightarrow V = \left( {\dfrac{2}{{R + 5\omega }}} \right)0.6 \times 5\omega \cr
  & \therefore V = \left( {\dfrac{{6\omega }}{{R + 5\omega }}} \right) \cr} $
This potential difference will be equal to source emf which is balanced. So
$\eqalign{
  & V = \left( {\dfrac{{6\omega }}{{R + 5\omega }}} \right) = 6mV \cr
  & \Rightarrow \left( {\dfrac{\omega }{{R + 5\omega }}} \right) = 1 \times {10^{ - 3}}V \cr
  & \Rightarrow 1000\omega = R + 5\omega \cr
  & \therefore R = 995\omega \cr} $

Hence option B will be the answer.

Note:
Here balancing the source emf means, at the balancing point if we place an ammeter between the source cell and potentiometer wire, no current will be passed through that. But internally current will be flowing in the potentiometer circuit. There will be no current between the potentiometer and source emf. Same principle can be used to find the internal resistance of the cell.