Question

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

• A. $k(l_2-l_1)$ and $kl_2$
• B. $kl_1$ and $k(l_2-l_1)\,$
• C. $k(l_2-l_1)$ and $kl_1$
• D. $kl_1$ and $kl_2$

Answer 1

Answer: (B)

$kl_1$ and $k(l_2-l_1)\,$

Answer 2

When the two way key is switched off, then
The current flowing in the resistors $R$ and $X$ is $\, \, \, \, I = 1\, A$...(i)
When the key between the terminals $1$ and $2$ is plugged in, then
Potential difference across $R = IR = kl_1$..(ii)
where $k$ is the potential gradient across the potentiometer wire
When the key between the terminals $1$ and $3$ is plugged in, then
Potential difference across $(R+X)=I (R+X)=kl_2$ ...(iii)
From equation (ii), we get
$\, \, \, \, R = \frac{kl_1}{I} =\frac{kl_1}{I}= kl_1\, \Omega$...(iv)
From equation (iii), we get
$\, \, \, \, R +X = \frac{kl_2}{I} =\frac{kl_2}{I}= kl_2\, \Omega$ (Using(i))
$X = kl_2 - R$
$= kl_2 - kl_1$ (Using(iv)
$= k(l_2 - l_1)\, \Omega$