
Two cells when connected in series are balanced on 8 m on a potentiometer. If the cells are connected with polarities of one of the cells reversed, then they balance on 2 m. The ratio of emf’s of two cells is
A. 3:4
B. 4:3
C. 3:5
D. 5:3
Answer
577.2k+ views
Hint: In a potentiometer, the net emf is the sum of individual emf when they are connected in series and it is the difference of individual emf when they are connected in parallel.
Complete step by step answer:
A potentiometer is a three terminal resistor with a sliding or rotating contact that forms an adjustable voltage divider. When only two terminals are used, it acts as a variable resistor or rheostat. It consists of a long wire of uniform area which is of 10 m length. The material of the wire should have a high resistivity and low temperature. The two wires are joined parallel to each other. It is an instrument that can be used to measure the emf or the internal resistance of an unknown source. It also has a number of other useful applications.
The potentiometer works on the principle that when a constant current flows through a wire of uniform cross section area, potential difference between its two points is directly proportional to the length of the wire between two points i.e., $emf = kl$ where k is a constant and l is the length of the wire at balanced point.
When two cells are connected in series, total emf is sum of the individual emf’s i.e., ${E_1} + {E_2} = k{l_1} + k{l_2}$
Total emf $\left( {{E_s}} \right) = k\left( {{l_1} + {l_2}} \right)$
${E_s} = k\left( {8 + 2} \right)$
$\implies {E_s} = 10k$
When the cells are connected in parallel (on changing their polarities), total emf is difference of the individual emf’s i.e.,${E_1} - {E_2} = k{l_1} - k{l_2}$
Total emf ${E_p} = k{l_1} - k{l_2}$
${E_p} = k\left( {{l_1} - {l_2}} \right)$
$\implies {E_p} = k\left( {8 - 2} \right)$
$\implies {E_p} = 6k$
Ratio of the two emf’s = $\dfrac{{{E_s}}}{{{E_p}}}$
$ \therefore \dfrac{{10k}}{{6k}} = \dfrac{5}{3}$
So, the correct answer is “Option D”.
Note:
When the cells are connected in series, the potential difference increases i.e., emf increases and when they are connected in parallel, the potential difference decreases.
Complete step by step answer:
A potentiometer is a three terminal resistor with a sliding or rotating contact that forms an adjustable voltage divider. When only two terminals are used, it acts as a variable resistor or rheostat. It consists of a long wire of uniform area which is of 10 m length. The material of the wire should have a high resistivity and low temperature. The two wires are joined parallel to each other. It is an instrument that can be used to measure the emf or the internal resistance of an unknown source. It also has a number of other useful applications.
The potentiometer works on the principle that when a constant current flows through a wire of uniform cross section area, potential difference between its two points is directly proportional to the length of the wire between two points i.e., $emf = kl$ where k is a constant and l is the length of the wire at balanced point.
When two cells are connected in series, total emf is sum of the individual emf’s i.e., ${E_1} + {E_2} = k{l_1} + k{l_2}$
Total emf $\left( {{E_s}} \right) = k\left( {{l_1} + {l_2}} \right)$
${E_s} = k\left( {8 + 2} \right)$
$\implies {E_s} = 10k$
When the cells are connected in parallel (on changing their polarities), total emf is difference of the individual emf’s i.e.,${E_1} - {E_2} = k{l_1} - k{l_2}$
Total emf ${E_p} = k{l_1} - k{l_2}$
${E_p} = k\left( {{l_1} - {l_2}} \right)$
$\implies {E_p} = k\left( {8 - 2} \right)$
$\implies {E_p} = 6k$
Ratio of the two emf’s = $\dfrac{{{E_s}}}{{{E_p}}}$
$ \therefore \dfrac{{10k}}{{6k}} = \dfrac{5}{3}$
So, the correct answer is “Option D”.
Note:
When the cells are connected in series, the potential difference increases i.e., emf increases and when they are connected in parallel, the potential difference decreases.
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