Question:
In the given figure, the emf of the cell is $2.2 \mathrm{~V}$ and if internal resistance is $0.6 \Omega$. Calculate the power dissipated in the whole circuit :
Correct Option: , 3
Solution:
$\frac{1}{R_{e q}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{6}=\frac{6+3+2+4}{24}=\frac{15}{24}$
$\mathrm{R}_{\mathrm{eq}}=\frac{24}{15}=1.6 \Rightarrow \mathrm{R}_{\mathrm{T}}=1.6+0.6=2.2 \Omega$
$\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{T}}}=\frac{(2.2)^{2}}{2.2}=2.2 \mathrm{~W}$
Option (3)
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