Question

In a potentiometer wire experiment, the emf of a battery in the primary circuit is $$20V$$ and its internal resistance is $$5\mathrm{\Omega }$$ . There is a resistance box in series with the battery and the potentiometer wire, whose resistance can be varied from $$120\mathrm{\Omega }$$ to $$170\mathrm{\Omega }$$ . Resistance of the potentiometer wire is $$75\mathrm{\Omega }$$ . The following potential differences can be measured using this potentiometer
$$(A)5V$$
$(B)6V$
$(C)7V$
$(D)8V$

Answer 1

Hint: We can solve this problem by finding the current value that flows through the circuit. The value of the current can be found by the values of electromotive force of the and the resistances of the potentiometer.

Formula used:
$I={\displaystyle \frac{E}{R+r+R_l}}$,
 where,$I$ is the current flow across the circuit, $E$ is the emf of the potentiometer,
$R$ is the Resistance box in series,$r$ Internal resistance of the cell, $R_i$ is the Resistance of the potentiometer wire.

Complete step by step answer:
 Given details:
emf of the cell, $$E=20V$$
Internal resistance, $$r=5\mathrm{\Omega }$$
Resistance box, $$R=120\mathrm{\Omega }$$to$$170\mathrm{\Omega }$$,
Resistance of the potentiometer, $$R_l=75\mathrm{\Omega }$$
Now considering the resistance, $$R=120\mathrm{\Omega }$$ the current produced is equal to
$I={\displaystyle \frac{E}{R+r+R_l}}$
Using the given values,
$\Rightarrow I={\displaystyle \frac{20}{120+5+75}}$
$\Rightarrow I={\displaystyle \frac{20}{200}}$
$\Rightarrow I=0.1A$
Hence the voltage that can be measured is $V=I{R}_i$
$V=0.1\times 75$
$\Rightarrow V=7.5V$
And when, $$R=170\mathrm{\Omega }$$ the current measured is $I={\displaystyle \frac{E}{R+r+R_l}}$
Using the given values,
$\Rightarrow I={\displaystyle \frac{20}{170+5+75}}$
$\Rightarrow I={\displaystyle \frac{20}{250}}$
$\Rightarrow I=0.08A$
Hence the voltage measured is $V=I{R}_i$
Substituting the values of the current and resistance in the formula.
$V=0.08\times 75$
$\Rightarrow V=6V$
So, the potential difference varies from $6V$ to $$7.5V$$.

Hence the right answer in the options $(B)$ and $(C)$.

Note:
1. The emf of a cell: The electromotive force of a cell is the maximum potential difference between two electrodes of a cell.
2. A potentiometer is used to measure the potential difference in a circuit and it has limited bandwidth.
3. The major disadvantage of potentiometers is that it requires a large force to move their sliding contacts.
4. If only two terminals potentiometer are used, one end and the wiper, it acts as a variable resistor or rheostat.The internal resistance of a cell refers to the opposition to the flow of current offered by the cells and the batteries itself generating heat, it is measured in $$ohm\left(\mathrm{\Omega }\right).$$