Consider the four circuits shown below. In wh | vedclass

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

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(A) The power dissipated in a circuit is given by $P = \frac{E^2}{R_{eq}}$,where $E$ is the $EMF$ of the source and $R_{eq}$ is the equivalent resistance of the circuit. To maximize power $P$,we need to minimize the equivalent resistance $R_{eq}$.For circuit $A$: Two resistors $R$ are in parallel. $R_{eq,A} = \frac{R \cdot R}{R + R} = \frac{R}{2} = 0.5R$.For circuit $B$: Two resistors $R$ are in series. $R_{eq,B} = R + R = 2R$.For circuit $C$: Two resistors $R$ are in parallel,and this combination is in series with another resistor $R$. $R_{eq,C} = \frac{R}{2} + R = 1.5R$.For circuit $D$: Two resistors $R$ are in series,and this combination is in parallel with another resistor $R$. $R_{eq,D} = \frac{(2R) \cdot R}{2R + R} = \frac{2R^2}{3R} = \frac{2}{3}R \approx 0.67R$.Comparing the equivalent resistances: $R_{eq,A} (0.5R) Since circuit $A$ has the minimum equivalent resistance,it will dissipate the maximum power.

Consider the four circuits shown below. In which circuit is the power dissipated the maximum? (Neglect the internal resistance of the applied power source $E$)

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