Question

In the given potentiometer circuit, for key K₁ closed, null point is at J₁. For K₂ closed, null point is at J₂. The value of $\frac{E_1}{E_2}$ is $\frac{a}{b}$, where a= _________ . (Note: Using standard balancing lengths $l_1$ and $l_2$ typically provided in this diagram, e.g., $l_1=4$m, $l_2=2$m).

Hint:

A potentiometer measures the EMF of a cell without drawing any current from it, which is why it is more accurate than a voltmeter.

Answer 1

Correct Answer: 1

Step 1: In a potentiometer, the potential $E$ is proportional to the balancing length $l$.
Step 2: $E_1 = \phi l_1$ and $E_2 = \phi l_2$, where $\phi$ is the potential gradient.
Step 3: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.
Step 4: Based on the ratios in such problems, if $l_1 = 2 l_2$, then $a/b = 2/1$.
Step 5: $a = 2$.