JEE Mains · Physics · STD 12 - 3. current electricity
Two cells are connected between points \(A\) and \(B\) as shown. Cell \(1\) has emf of \(12\,V\) and internal resistance of \(3\,\Omega\). Cell \(2\) has emf of \(6\,V\) and internal resistance of \(6\,\Omega\). An external resistor \(R\) of \(4\,\Omega\) is connected across \(A\) and \(B\). The current flowing through \(R\) will be \(.............A\).
- A \(0\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(E _{ eq }=\frac{\frac{12}{3}-\frac{6}{6}}{\frac{1}{3}+\frac{1}{6}}\) \(E _{ eq }=6\,V\) \(r _{ eq }=2\,\Omega\) \(R =4\,\Omega\) So, \(i=\frac{6}{2+4}=1\,A\)
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