Question

An ideal battery of $4V$ and resistance $R$ are connected in series in the primary circuit of a potentiometer of length $1m$ and resistance $5\Omega$. The value of $R$, to given a potential difference of $5mV$ across $10cm$ of potentiometer wire is:
A) $490\Omega$
(B) $480\Omega$
(C) $395\Omega$
(D) $495\Omega$

Answer 1

Recall the potential gradient, its definition and formula. The resistance for $1m$ length potentiometer is given, so find the resistance of the potentiometer wire then find out the value of the resistance connected in series with the battery.

Formula Used:
$i = \dfrac{V}{R}$
Where $i$ is the current flowing
$V$ is the voltage difference
$R$ is the resistance

Complete step by step answer:
In the question we have given the voltage of the ideal battery, that is
$V = 4V$
And the battery and a resistance is connected in the series, so the equivalent resistance become
${R_{eq}} = R + 5$
So, the current flowing, $i = \dfrac{V}{{R + 5}}$
$\Rightarrow i = \dfrac{4}{{R + 5}}A$
Now, in the question we have given the resistance $5\Omega$, for $1m$ length of potentiometer
So, for $10cm$ length of potentiometer, the resistance becomes
$R' = 5 \times \dfrac{{10}}{{100}}$
$\Rightarrow R' = 0.5\Omega$
We have also given the potential difference across $10cm$ of potentiometer wire, that is
$\Delta V' = 5mV$
Now, we have to convert it into the SI unit that is volt.
$\Rightarrow \Delta V' = 5 \times {10^{ - 3}}V$
Now apply the formula for ohm's law,
$\Delta V' = iR'$
On putting the values of all the available variables, we get
$\Rightarrow 5 \times {10^{ - 3}} = \left( {\dfrac{4}{{R + 5}}} \right) \times 0.5$
$\Rightarrow \dfrac{4}{{R + 5}} = 10$
On further solving,
$\Rightarrow R + 5 = 400$
Finally we get the value of the R,
$R = 395\Omega$
Thus, the value of resistance $R$ connected in series is given by $395\Omega$.
Therefore, the correct answer is option (C).

Note: Potential gradient is defined as the change in potential difference with respect to the per unit length. The potentiometer is a three terminal variable resistor in which the resistance is manually varied to control the flow of electric current. It acts as an adjustable voltage divider. The potentiometer problems are similar to Wheatstone bridge problems. Balancing the resistance is what you do to find the desired quantity.