The maximum power dissipated in an external r | vedclass

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(D) The power $P$ dissipated in an external resistance $R$ connected to a cell of emf $\varepsilon$ and internal resistance $r$ is given by the formul

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$\frac{\varepsilon^2}{4r}$

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(D) The power $P$ dissipated in an external resistance $R$ connected to a cell of emf $\varepsilon$ and internal resistance $r$ is given by the formula:$P = I^2 R = \left( \frac{\varepsilon}{R+r} ight)^2 R$To find the maximum power,we differentiate $P$ with respect to $R$ and set it to zero:$\frac{dP}{dR} = \varepsilon^2 \left[ \frac{(R+r)^2(1) - R(2)(R+r)}{(R+r)^4} ight] = 0$$(R+r)^2 - 2R(R+r) = 0$$(R+r)(R+r - 2R) = 0$$r - R = 0 \implies R = r$Thus,the power dissipated is maximum when the external resistance equals the internal resistance $(R = r)$.Substituting $R = r$ into the power equation:$P_{\max} = \left( \frac{\varepsilon}{r+r} ight)^2 r = \left( \frac{\varepsilon}{2r} ight)^2 r = \frac{\varepsilon^2}{4r^2} 	imes r = \frac{\varepsilon^2}{4r}$

The maximum power dissipated in an external resistance $R$,when connected to a cell of emf $\varepsilon$ and internal resistance $r$,will be . . . . . . .

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