Question 496: Current Electricity

Question

When cell of e.m.f. ' $E_1$ ' is connected to potentiometer wire, the balancing length is ' $l_l$ '. Another cell of e.m.f. ' $E_2$ ' $\left(E_1>E_2\right)$ is connected so that two cells oppose each other, the balancing length is ' $l_1$ '. The ratio $E_1: E_2$ is

Accepted Answer

Step 1: Potentiometer principle
In potentiometer experiments,
$ 	ext{E.M.F.} \propto 	ext{balancing length}. $
So if a cell of emf $E$ alone is connected, balancing length $l \propto E$.
Thus,
$ E_1 \;\propto\; l_1. $
Step 2: When cells oppose
If another cell $E_2$ is connected in
opposition
, the effective emf across the potentiometer = $E_1 - E_2$.
Its balancing length should be proportional:
$ E_1 - E_2 \;\propto\; l_2. $
Step 3: Ratio relations
From step 1:
$\frac{E_1}{E_1-E_2} = \frac{l_1}{l_2}$.
Ã¢ÂÂ $E_1 l_2 = (E_1 - E_2) l_1$.
Ã¢ÂÂ $E_1 l_2 = E_1 l_1 - E_2 l_1$.
Ã¢ÂÂ $E_2 l_1 = E_1(l_1 - l_2)$.
Ã¢ÂÂ $\frac{E_1}{E_2} = \frac{l_1}{\,l_1 - l_2}$.
Step 4: Final Answer
So the ratio is
$ \boxed{\;\;\dfrac{E_1}{E_2} \;=\;\dfrac{l_1}{l_1 - l_2}\;} $
Correct Option: (B) $	frac{l_1}{l_1 - l_2}$.
Ã¢ÂÂ

Answer

$\frac{l_1}{l_1+l_2}$

Answer

$\frac{l_1}{l_1-l_2}$

Answer

$\frac{l_1-l_2}{l_1}$

Answer

$\frac{l_1+l_2}{l_1-l_2}$

Current Electricity

When cell of e.m.f. ' E_1 ' is connected to potentiometer wire, the balancing length is ' l_l '. Another cell of e.m.f. ' E_2 ' \left(E_1>E_2\right) is connected so that two cells oppose each other, the balancing length is ' l_1 '. The ratio E_1: E_2 is

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