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JEE Mains · Physics · STD 12 - 3. current electricity
A \(dc\) source of \(emf \,E_1 = 100\,V\) and internal resistance \(r = 0.5\,\Omega ,\) a storage battery of emf \(E_2 = 90\,V\) and an external resistance \(R\) are connected as shown in figure. For what value of \(R\) no current will pass through the battery ? ................ \(\Omega\)
- A \(5.5\)
- B \(3.5\)
- C \(4.5\)
- D \(2.5\)
Answer & Solution
Correct Answer
(C) \(4.5\)
Step-by-step Solution
Detailed explanation
\(\frac{100}{R+r}=\frac{90}{R}\) \(\Rightarrow \frac{R+r}{R}=\frac{10}{9}\) \(\Rightarrow \quad 1+\frac{0.5}{\mathrm{R}}=\frac{10}{9}\) \(\Rightarrow \quad \frac{0.5}{\mathrm{R}}=\frac{1}{9}\) \(\therefore R=4.5 \,\Omega\)
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