Consider the four circuits shown in the figur | vedclass

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(A) The power dissipated in a circuit is given by $P = \frac{\varepsilon^2}{R_{eq}}$,where $\varepsilon$ is the $EMF$ of the battery and $R_{eq}$ is t

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(A) The power dissipated in a circuit is given by $P = \frac{\varepsilon^2}{R_{eq}}$,where $\varepsilon$ is the $EMF$ of the battery and $R_{eq}$ is the equivalent resistance. To maximize power $P$,we must minimize the equivalent resistance $R_{eq}$.For circuit $A$: $R_{eq} = \frac{R 	imes R}{R + R} = \frac{R}{2} = 0.5R$.For circuit $B$: $R_{eq} = R + R = 2R$.For circuit $C$: $R_{eq} = \frac{R}{2} + R = 1.5R$.For circuit $D$: $R_{eq} = \frac{(2R) 	imes R}{2R + R} = \frac{2R^2}{3R} = \frac{2}{3}R \approx 0.67R$.Comparing the values of $R_{eq}$,circuit $A$ has the minimum equivalent resistance $(0.5R)$. Therefore,the power dissipated is the greatest in circuit $A$.

Consider the four circuits shown in the figure given below. In which circuit is the power dissipated the greatest? (Neglect the internal resistance of the power supply)

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