A battery of e.m.f 3.0V is connected across a potentiometer wire AB of length 10\(\Omega m^{-1}\) as illustrated in the diagram above. If the jockey J makes contact at the 30cm mark, determine the resistance of length AJ and voltage across AJ.
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Correct Answer: Option A
Explanation:
E = 3.0V; \(\frac{R}{l} = 10\Omega m^{-1}\)
l = 30cm = 0.3m; R = \(10\Omega m^{-1} \times 0.3m = 3\Omega\)
\(\frac{E}{V} = \frac{l_{1}}{l_{2}} \implies \frac{3}{V} = \frac{1}{0.3}\)
= \(\frac{3\times 0.3}{1} = 0.9V\)
E = 3.0V; \(\frac{R}{l} = 10\Omega m^{-1}\)
l = 30cm = 0.3m; R = \(10\Omega m^{-1} \times 0.3m = 3\Omega\)
\(\frac{E}{V} = \frac{l_{1}}{l_{2}} \implies \frac{3}{V} = \frac{1}{0.3}\)
= \(\frac{3\times 0.3}{1} = 0.9V\)