JEE Mains · Physics · STD 12 - 3. current electricity
Two cells of same emf but different internal resistances \(I_{1}\) and \(I_{2}\) are connected in series with a resistance \(R\). The value of resistance \(R\), for which the potential difference across second cell is zero, is
- A \(r _{2}- r _{1}\)
- B \(r_{1}-r_{2}\)
- C \(r _{1}\)
- D \(r _{2}\)
Answer & Solution
Correct Answer
(A) \(r _{2}- r _{1}\)
Step-by-step Solution
Detailed explanation
\(I=\frac{2 E }{R+I_{1}+I_{2}} \ldots \text{(i)}\) But \(V _{ A }- V _{ B }= E - I _{2}=0\) \(\Rightarrow I =\frac{ E }{ I _{2}} \ldots \text{(ii)}\) Comparing values of \(I\) from \((i)\) and \((ii)\) \(\frac{E}{r_{2}}=\frac{2 E}{R+r_{1}+r_{2}}\) \(\Rightarrow R=r_{2}-r_{1}\)
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