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Power dissipated across the \(8 \Omega\) resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the \(3 \Omega\) resistor is
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Power \(=\mathrm{V} . \mathrm{I}=\mathrm{I}^2 \mathrm{R}\)
\(\mathrm{i}_2=\sqrt{\frac{\text { Power }}{\mathrm{R}}}=\sqrt{\frac{2}{8}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
Potential over \(8 \Omega=\mathrm{Ri}_2=8 \times \frac{1}{2}=4 \mathrm{~V}\) This is the potential over parallel branch. So, \(i_1=\frac{4}{4}=1\)
Power of \(3 \Omega=\mathrm{i}_1{ }^2 \mathrm{R}=1 \times 1 \times 3=3 \mathrm{~W}\)
\(\mathrm{i}_2=\sqrt{\frac{\text { Power }}{\mathrm{R}}}=\sqrt{\frac{2}{8}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
Potential over \(8 \Omega=\mathrm{Ri}_2=8 \times \frac{1}{2}=4 \mathrm{~V}\) This is the potential over parallel branch. So, \(i_1=\frac{4}{4}=1\)
Power of \(3 \Omega=\mathrm{i}_1{ }^2 \mathrm{R}=1 \times 1 \times 3=3 \mathrm{~W}\)
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