Using the principle of potentiometer, how do you compare the potential of two primary cells?
Answer
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Hint:A potentiometer is a device that is used to measure the potential difference between two points. The potentiometer works on the principle that when a constant current flows through a wire of uniform cross sectional area, potential difference(E) between its two points is directly proportional to the length(l) of the wire between the two points, \[E \propto l\] .
Complete answer:
To compare the potential of two primary cell using a potentiometer we use the following setup. Connect the positive pole of the battery (a battery of constant e.m.f.) to the zero end (A) of the potentiometer and the negative pole through a one-way key(K1), an ammeter and a low resistance rheostat to the other end (C) of the potentiometer. The end of the potentiometer is connected to two primary cells(${E_1}$and \[{E_2}\]). These two primary cells are connected to a galvanometer via a two way key. The galvanometer is connected to a high resistance box and then to the Jockey. The Jockey has to find the point of null deflection in the galvanometer(${N_1}$and${N_2}$). Now we take out a resistance from the resistance box and simultaneously check for the length of the wire giving us the null point using the jockey (sliding the Jockey over the wire) by connecting each of the primary cells at a time with the two way key.
Let the balancing length for the first primary cell ${l_1}$ and for the second primary cell be ${l_2}$.
Now using the principle of potentiometer we can write,
$
{E_1} \propto {l_1} \\
\Rightarrow {E_1} = k{l_1} \\ $ ------------(1)
$
{E_2} \propto {l_2} \\
\Rightarrow {E_2} = k{l_2} \\ $ ------------(2)
Dividing both the equations we get,
\[ \therefore \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{l_1}}}{{{l_2}}}\]
Note:Using the principle of potentiometer we can say that the potential of the two primary cells is the ratio of the balancing length of the potentiometer (the length that gives the null point or zero deflection galvanometer). The smallest amount of potential difference that can be measured by a potentiometer is called the sensitivity of the potentiometer, it can be increased by increasing the length of the potentiometer wire or decreasing the current flowing through it.
Complete answer:
To compare the potential of two primary cell using a potentiometer we use the following setup. Connect the positive pole of the battery (a battery of constant e.m.f.) to the zero end (A) of the potentiometer and the negative pole through a one-way key(K1), an ammeter and a low resistance rheostat to the other end (C) of the potentiometer. The end of the potentiometer is connected to two primary cells(${E_1}$and \[{E_2}\]). These two primary cells are connected to a galvanometer via a two way key. The galvanometer is connected to a high resistance box and then to the Jockey. The Jockey has to find the point of null deflection in the galvanometer(${N_1}$and${N_2}$). Now we take out a resistance from the resistance box and simultaneously check for the length of the wire giving us the null point using the jockey (sliding the Jockey over the wire) by connecting each of the primary cells at a time with the two way key.
Let the balancing length for the first primary cell ${l_1}$ and for the second primary cell be ${l_2}$.
Now using the principle of potentiometer we can write,
$
{E_1} \propto {l_1} \\
\Rightarrow {E_1} = k{l_1} \\ $ ------------(1)
$
{E_2} \propto {l_2} \\
\Rightarrow {E_2} = k{l_2} \\ $ ------------(2)
Dividing both the equations we get,
\[ \therefore \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{l_1}}}{{{l_2}}}\]
Note:Using the principle of potentiometer we can say that the potential of the two primary cells is the ratio of the balancing length of the potentiometer (the length that gives the null point or zero deflection galvanometer). The smallest amount of potential difference that can be measured by a potentiometer is called the sensitivity of the potentiometer, it can be increased by increasing the length of the potentiometer wire or decreasing the current flowing through it.
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