JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2021
- Hard
- Numerical
First, a set of \({n}\) equal resistors of \(10\; \Omega\) each are connected in series to a battery of emf \(20\; {V}\) and internal resistance \(10\; \Omega .\) A current \(I\) is observed to flow. Then, the \(n\) resistors are connected in parallel to the same battery. It is observed that the current is increased \(20\) times, then the value of \(n\) is .... .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(20\)
Step-by-step Solution
Detailed explanation
In series
\({R}_{{eq}}={nR}=10 {n}\)
\({i}_{{s}}=\frac{20}{10+10 {n}}=\frac{2}{1+{n}}\)
in parallel
\({R}_{{eq}}=\frac{10}{{n}}\)
\({i}_{{p}}=\frac{20}{\frac{10}{{n}}+10}=\frac{2 {n}}{1+{n}}\)
\(\frac{{i}_{{p}}}{{i}_{{s}}}=20\)
\(\frac{\left(\frac{2 {n}}{1+{n}}\right)}{\left(\frac{2}{1+{n}}\right)}=20\)
\({n}=20\)
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