Current Electricity - Result Question 101
106. A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is $k$ volt $/ cm$ and the ammeter, present in the circuit, reads $1.0 A$ when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths $l_1 cm$ and $l_2 cm$ respectively. The magnitudes, of the resistors $R$ and $X$, in ohms, are then, equal, respectively, to
[2010]
(a) $k(l_2-l_1)$ and $k l_2$
(b) $k l_1$ and $k(l_2-l_1)$
(c) $k(l_2-l_1)$ and $k l_1$
(d) $k l_1$ and $k l_2$
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Answer:
Correct Answer: 106. (b)
Solution:
- (b) (i) When key between the terminals 1 and 2 is plugged in,
P.D. across $R=I R=k l_1$
$\Rightarrow R=k l_1$ as $I=1 A$
(ii) When key between terminals 1 and 3 is plugged in,
P.D. $across(X+R)=I(X+R)=k l_2$
$\Rightarrow X+R=k l_2$
$\therefore \quad X=k(l_2-l_1)$
$\therefore \quad R=k l_1$ and $X=k(l_2-l_1)$
