In the circuit shown in figure, the power whi | vedclass

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Question

Current in $6\, \Omega \quad \mathrm{I}=\sqrt{\frac{\mathrm{P}}{\mathrm{R}}}=\sqrt{\frac{6}{6}}=1$

${\mathrm{I}=\frac{12}{6+\frac{8 \mathrm{P}}{8+\

Vedclass · Accepted Answer

$24$

Vedclass · Answer

Current in $6\, \Omega \quad \mathrm{I}=\sqrt{\frac{\mathrm{P}}{\mathrm{R}}}=\sqrt{\frac{6}{6}}=1$

${\mathrm{I}=\frac{12}{6+\frac{8 \mathrm{P}}{8+\mathrm{R}}}} $

${1=\frac{12}{48+14 \mathrm{R}}(8+\mathrm{R})}$

${\mathrm{R}=24\, \Omega}$

In the circuit shown in figure, the power which is dissipated as heat in the $6\,\Omega $ resistor is $6\,W$. What is the value of resistance $R$ in the circuit? ................ $\Omega$

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