JEE MAIN - Physics (2002 - No. 8)
If in the circuit, power dissipation is $$150W,$$ then $$R$$ is
$$2\,\Omega $$
$$6\,\Omega $$
$$5\,\Omega $$
$$4\,\Omega $$
Explanation
The equivalent resistance is $${{\mathop{\rm R}\nolimits} _{eq}} = {{2 \times R} \over {2 + R}}$$
$$\therefore$$ Powder dissipation $$P = {{{V^2}} \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$$
$$\therefore$$ $$150 = {{15 \times 15} \over {{R_{eq}}}}$$
$$\therefore$$ $${{\mathop{\rm R}\nolimits} _{eq}} = {{15} \over {10}} = {3 \over 2}$$
$$ \Rightarrow {{2R} \over {2 + R}} = {3 \over 2}$$
$$ \Rightarrow 4R = 6 + 3R$$
$$ \Rightarrow R = 6\Omega $$
$$\therefore$$ Powder dissipation $$P = {{{V^2}} \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$$
$$\therefore$$ $$150 = {{15 \times 15} \over {{R_{eq}}}}$$
$$\therefore$$ $${{\mathop{\rm R}\nolimits} _{eq}} = {{15} \over {10}} = {3 \over 2}$$
$$ \Rightarrow {{2R} \over {2 + R}} = {3 \over 2}$$
$$ \Rightarrow 4R = 6 + 3R$$
$$ \Rightarrow R = 6\Omega $$
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