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  • In the given circuit of the potentiometer, the potential difference E across AB (10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at J1</su

In the given circuit of the potentiometer, the potential difference E across AB (10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at J1 so that there is no deflection in the galavanometer. Now the first battery (E1) is replaced by the second battery (E2) for working by making K1 open and K2 closed. The galvanometer gives then null deflection at J2. The value of \frac{E_{1}}{E_2} is \frac{a}{b}where a = ______
Option: 1 1
Option: 2 2
Option: 3 3
Option: 4 4

Answers (1)

best_answer

Length of  AB=10 m
For battery \mathrm{E}_{1}, balancing length is \mathrm{l}_{1} and l_{1}=380 \mathrm{~cm} [from end A]
For battery \mathrm{E}_{2}, balancing length is \mathrm{l}_{2} and  l_{1}=760 \mathrm{~cm} [from end A]

Now, we know that

\\ \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}} \\ \\ \Rightarrow \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{380}{760}=\frac{1}{2}=\frac{\mathrm{a}}{\mathrm{b}}
\\ \therefore a=1 \ \& \ b=2 \\ \Rightarrow \mathrm{a}=1

Posted by

avinash.dongre

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