The resistance $R_{1}$ and $R_{2}$ are joined in parallel and a current is passed so that the amount of heat liberated is $H_{1}$ and $H_{2}$ respectively. The ratio $\frac{H_{1}}{H_{2}}$ has the value

\frac{R_{2}}{R_{1}}

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\frac{R_{1}}{R_{2}}

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\frac{R_{1}^{2}}{R_{2}^{2}}

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\frac{R_{2}^{2}}{R_{1}^{2}}

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Solution

The correct option is B $\frac{R_{2}}{R_{1}}$
Heat produced, $H = \frac{V^{2} t}{R}$
both resistances are connected in parallel combination that's why both will get equal voltage drop hence heat produced will directly depend upon resitance value of them.
i.e. $H \propto \frac{1}{R}$

so $\frac{H_{1}}{H_{2}} = \frac{R_{2}}{R_{1}}$

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Similar questions

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

If the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle of radius R then

(a) $R_{1} + R_{2} = R$

(b) $R_{1} + R_{2} > R$

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

R_{1}

R_{2}

R_{1} + R_{2} = R

R_{1}^{2} + R_{2}^{2} = R^{2}

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R_{2}

r_{1}

r_{2}

R

M

F_{1}

F_{2}

\frac{{R_{1}}^{2}}{{R_{2}}^{2}} = \frac{9}{576}, \frac{R_{1}}{R_{2}} = ?

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