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 $1.0A$ 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 $k{l}_2$
• 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 $k{l}_2$

Answer 1

Answer: (B)

When the two way key is switched off, then
The current flowing in the resistors $R$ and $X$ is $I=1A$...(i)
When the key between the terminals $1$ and $2$ is plugged in, then
Potential difference across $R=IR=k{l}_1$..(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 $(R+X)=I(R+X)=k{l}_2$ ...(iii)
From equation (ii), we get
$R=\frac{k{l}_1}{I}=\frac{k{l}_1}{I}=k{l}_1\Omega$...(iv)
From equation (iii), we get
$R+X=\frac{k{l}_2}{I}=\frac{k{l}_2}{I}=k{l}_2\Omega$ (Using(i))
$X=k{l}_2-R$
$=k{l}_2-k{l}_1$ (Using(iv)
$=k(l_2-l_1)\Omega$