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 $W$)
JEE MAIN 2021, Diffcult
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$\frac{1}{R_{e q}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{6}=\frac{6+3+2+4}{24}=\frac{15}{24}$
$R_{e q}=\frac{24}{15}=1.6 \Rightarrow R_{T}=1.6+0.6=2.2 \,\Omega$
$P=\frac{V^{2}}{R_{T}}=\frac{(2.2)^{2}}{2.2}=2.2 \,{W}$
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