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

The Correct Option is B

To determine the values of resistors \( R \) and \( X \) using a potentiometer setup, we need to understand the relationship between potential difference, potential gradient, and balance point lengths.

The potential difference \( (V) \) across the length of the potentiometer wire is given by the formula:

V = k \cdot l

where \( k \) is the potential gradient in volts per cm, and \( l \) is the balance length in cm.

Given that the ammeter reads 1.0 A, it indicates the current flowing through the circuit when the two-way key is switched off.

Let's analyze the two cases:

1. When the key is plugged between terminals 1 and 2: The potential difference across the resistor \( R \) will balance at length \( l_1 \). Therefore, the voltage across \( R \) is: V_R = k \cdot l_1
2. When the key is plugged between terminals 1 and 3: The combined potential difference across resistors \( R + X \) will balance at length \( l_2 \). Thus, the total voltage is: V_{R+X} = k \cdot l_2

From these two cases, we can write the equations:

• V_R = k \cdot l_1\
• V_{R+X} = k \cdot l_2\

By considering the circuit, we observe that:

• V_X = V_{R+X} - V_R\
• V_X = k \cdot l_2 - k \cdot l_1 = k(l_2-l_1)\

Thus, the resistance values are:

• R = k \cdot l_1\
• X = k(l_2-l_1)\

Hence, the magnitudes of the resistors are:

• \( R = kl_1 \)
• \( X = k(l_2-l_1) \)

The correct answer is thus: \( kl_1 \) and \( k(l_2-l_1) \).