JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2021
- Hard
- MCQ
In the given figure, the \(emf\) of the cell is \(2.2\, {V}\) and if internal resistance is \(0.6\, \Omega\). Calculate the power dissipated in the whole circuit: (in \(W\))

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Answer & Solution
Correct Answer
(C) \(2.2\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{R_{e q}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{6}=\frac{6+3+2+4}{24}=\frac{15}{24}\)
\(R_{e q}=\frac{24}{15}=1.6 \Rightarrow R_{T}=1.6+0.6=2.2 \,\Omega\)
\(P=\frac{V^{2}}{R_{T}}=\frac{(2.2)^{2}}{2.2}=2.2 \,{W}\)
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