The null point of a potentiometer with a cell of emf $\varepsilon$ is obtained at a distance $l$ on the wire…
The null point of a potentiometer with a cell of emf $\varepsilon$ is obtained at a distance $l$ on the wire, then
- $\varepsilon \propto l$
- $\varepsilon \propto l^2$
- $\varepsilon \propto \frac{1}{l}$
- $\varepsilon \propto \frac{1}{l^2}$
Solution
As we know, electric field, $E=-\frac{d V}{d r}$
Hence, emf $\varepsilon$ is also directly proportional to $l$ i.e., $\varepsilon \propto l$.
Asked in: AP EAMCET 2021 (25 Aug Shift 2)
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