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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2023
  • Medium
  • Numerical

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 \(.............A\).

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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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