Question

A potentiometer wire is $100cm$ long and a constant potential difference is maintained across it. Two cells are connected in series, first to support one another and then in the opposite direction. The balance points are obtained at $50cm$ and $10cm$ from the positive end of the wire in the two cases. The ratio of the emf’s is:
A. $5:1$
B. $5:4$
C. $3:4$
D. $3:2$

Answer 1

Hint: A potentiometer is a device that can be used to measure the emf of a cell. It works on the principle that the algebraic sum of all potential drops in a closed loop is zero, that is, Kirchhoff’s Voltage Law. This problem can be solved by taking the two balance points in consideration and finding out the voltage drops for the two cases. That will be the equivalent emfs of the combinations of the two cells in each case. From that emf of each cell can be found out.

Formula used:
$\sum{V=0\text{ }\left( \text{in a closed loop} \right)}$ ---(Kirchhoff’s Voltage Law)
$V\propto R$
where V is the potential difference across a resistor and R is the resistance of the resistor.
$R=\rho L$
where R is the resistance of the wire, $\rho $ is the resistance per unit length of the wire and L is the length of the wire.

Complete step by step answer:
A potentiometer, in essence, consists of a long wire of uniform resistance per unit length and a jockey. It makes use of Kirchhoff’s Voltage law to find out the emf of a cell. Kirchoff’s voltage law states that the algebraic sum of all potential drops in a closed loop is zero. That is,
$\sum{V=0\text{ }\left( \text{in a closed loop} \right)}$ ---(Kirchhoff’s Voltage Law)--(1)

By positioning the jockey on different points of the potentiometer wire the resistance in the circuit changes as
$R=\rho L$ --(2)
where R is the resistance of the wire, $\rho $ is the resistance per unit length of the wire and L is the length of the wire.
By placing the jockey at different positions L can be changed and effectively R changes.
This changes the potential drop in the circuit as
$V\propto R$ --(3)
where V is the potential difference across a resistor and R is the resistance of the resistor.
A cell is connected to two points on the potentiometer wire to form a complete circuit consisting of the cell and the part of the potentiometer wire. When we reach a balance point, potential drop across the part of the wire is effectively the emf of the cell since the sum of all potential drops will be zero in a closed loop.
This is how the potentiometer wire measures the emf of the cell.
Now let us analyze the question.
We are given two cells. Let the emfs of the two cells be ${{E}_{1}}$ and ${{E}_{2}}$ respectively.
The first case (Two cells supporting each other)-
Total emf of the cells $={{E}_{1}}+{{E}_{2}}$ (since emfs of cells in series in the same sense add up)--(4)
Length of balance point obtained $=50cm$
Using (2),
$\therefore R=50\rho $ where $\rho $ is the resistance per cm of the wire.
Now using (3)
Voltage drop across the wire $\left( V \right)=K50\rho $ --(5)
where K is a constant of proportionality.
Now, using (1),(4) and(5),
 ${{E}_{1}}+{{E}_{2}}=K50\rho $ --(6)
Now, The second case (Two cells opposing each other)-
Total emf of the cells $={{E}_{1}}-{{E}_{2}}$ (since emfs of cells in series in the opposite sense have to be subtracted. Here we consider ${{E}_{1}}>{{E}_{2}}$ )--(7)
Length of balance point obtained $=50cm$
Using (2),
$\therefore R=10\rho $ where $\rho $ is the resistance per cm of the wire.
Now using (8)
Voltage drop across the wire $\left( V \right)=K10\rho $ --(9)
where K is a constant of proportionality.
Now, using (1),(7) and(9),
 ${{E}_{1}}-{{E}_{2}}=K10\rho $ --(10)
Hence, adding (6) and (10), we get,
${{E}_{1}}+{{E}_{2}}+{{E}_{1}}-{{E}_{2}}=K50\rho +K10\rho $
$\therefore 2{{E}_{1}}=K60\rho $
$\therefore {{E}_{1}}=K30\rho $ --(11)
Subtracting (10) from (6), we get,
${{E}_{1}}+{{E}_{2}}-{{E}_{1}}+{{E}_{2}}=K50\rho -K10\rho $
$\therefore 2{{E}_{2}}=K40\rho $
$\therefore {{E}_{2}}=K20\rho $ --(12)
Therefore, comparing (11) and (12), to get the ratio, we get,
$\dfrac{{{E}_{1}}}{{{E}_{2}}}=\dfrac{K30\rho }{K20\rho }=\dfrac{3}{2}=3:2$
Hence, the required ratio is $3:2$.
Therefore, the correct option is D) $3:2$.

Note: In questions where the final answer is a ratio it is always better to assume constants of proportionality and only find the values for the necessary quantities. For example, in this question we considered $V\propto R$and not $V=IR$ as the common Ohm’s law is since at the end the current $I$ would have anyway cancelled out in the ratio. Moreover, we would have to find out the value of $I$ for each case which would have taken a longer time. The same goes for the constant $\rho $ in the equation for resistance. This saves a lot of time.
However, one must remember that this method is only useful when the answer is in the form of a ratio. If the question asks for a concrete value, we would have to get numerical values for the other parameters and not assume them as constants that do not affect our calculation.