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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
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A
$kl_1$ and $kl_2$
B
$k(l_2 - l_1)$ and $kl_2$
C
$kl_1$ and $k(l_2 - l_1)$
D
$k(l_2 - l_1)$ and $kl_1$
Detailed Solution
The current through R and X in the secondary circuit is I = 1.0 A.
With the key between 1 and 2, the potential difference across R alone is balanced: $V_R = IR = kl_1$
Since I = 1 A, $R = kl_1$ ohm
With the key between 1 and 3, the potential difference across R and X in series is balanced: $I(R + X) = kl_2$
$R + X = kl_2$, so $X = kl_2 - kl_1$
$X = k(l_2 - l_1)$ ohm
So R and X are $kl_1$ and $k(l_2 - l_1)$ respectively.
With the key between 1 and 2, the potential difference across R alone is balanced: $V_R = IR = kl_1$
Since I = 1 A, $R = kl_1$ ohm
With the key between 1 and 3, the potential difference across R and X in series is balanced: $I(R + X) = kl_2$
$R + X = kl_2$, so $X = kl_2 - kl_1$
$X = k(l_2 - l_1)$ ohm
So R and X are $kl_1$ and $k(l_2 - l_1)$ respectively.