MHT CET · Physics · Current Electricity
A cell can be balanced against \(110 \mathrm{~cm}\) and \(100 \mathrm{~cm}\) of potentiometer wire, respectively with and without being short circuited through a resistance of \(10 \Omega\). Its internal resistance is
- A \(1.0 \Omega\)
- B \(0.5 \Omega\)
- C \(2.0 \Omega\)
- D zero
Answer & Solution
Correct Answer
(A) \(1.0 \Omega\)
Step-by-step Solution
Detailed explanation
In potentiometer experiment in which we find internal resistance of a cell, let \(E\) be the emf of the cell and \(V\) the terminal potential difference, then
\(
\frac{E}{V}=\frac{\mathrm{h}}{l_{2}}
\)
where \(\mathrm{l}\) and \(l_{2}\) are lengths of potentiometer wire with and without short circuited through a resistance.
Since, \(\frac{E}{V}=\frac{R+r}{R}\)
\(\quad[\because E=I(R+r)\) and \(V=I R]\)
\(\therefore \quad \frac{R+r}{R}=\frac{4}{l_{2}}\)
or \(\quad 1+\frac{r}{R}=\frac{110}{100}\)
Or \(\frac{r}{R}=\frac{10}{100}\)
or \(r=\frac{1}{10} \times 10=1 \Omega\)
\(
\frac{E}{V}=\frac{\mathrm{h}}{l_{2}}
\)
where \(\mathrm{l}\) and \(l_{2}\) are lengths of potentiometer wire with and without short circuited through a resistance.
Since, \(\frac{E}{V}=\frac{R+r}{R}\)
\(\quad[\because E=I(R+r)\) and \(V=I R]\)
\(\therefore \quad \frac{R+r}{R}=\frac{4}{l_{2}}\)
or \(\quad 1+\frac{r}{R}=\frac{110}{100}\)
Or \(\frac{r}{R}=\frac{10}{100}\)
or \(r=\frac{1}{10} \times 10=1 \Omega\)
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