JEE Mains · Physics · STD 12 - 3. current electricity
Two identical cells each of emf \(1.5 \,V\) are connected in parallel across a parallel combination voltmeter connected in the circuit measures \(1.2 \,V\). The internal resistance of each cell is.................\(\Omega\)
- A \(2.5\)
- B \(4\)
- C \(5\)
- D \(10\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\(V = E - ir / 2\) \(1.2=1.5- i \left(\frac{ r }{2}\right)\) \(i \frac{ r }{2}=0.3\) \(i =\frac{1.5}{10+\frac{ r }{2}} \Rightarrow 10 i +\frac{ ir }{2}=1.5\) \(10 i =1.5-0.3\) \(i =0.12\, A\) \(\Rightarrow r =\frac{0.6}{0.12}=5 \,\Omega\)
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