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(D) Given,the resistance $R$ of a device is related to the current $I$ by the relation: $R = \frac{0.2 I}{I-4}$.Power $P$ consumed by the device is gi

Vedclass · Accepted Answer

$21.6$

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(D) Given,the resistance $R$ of a device is related to the current $I$ by the relation: $R = \frac{0.2 I}{I-4}$.Power $P$ consumed by the device is given by $P = I^2 R$.Substituting the expression for $R$: $P = I^2 \left( \frac{0.2 I}{I-4} ight) = \frac{0.2 I^3}{I-4}$.To find the minimum power,we differentiate $P$ with respect to $I$ and set it to zero: $\frac{dP}{dI} = 0$.$\frac{d}{dI} \left( \frac{0.2 I^3}{I-4} ight) = 0$.Using the quotient rule: $\frac{(I-4)(0.6 I^2) - (0.2 I^3)(1)}{(I-4)^2} = 0$.$0.6 I^3 - 2.4 I^2 - 0.2 I^3 = 0$.$0.4 I^3 - 2.4 I^2 = 0$.$0.4 I^2 (I - 6) = 0$.Since $I > 4$,we have $I = 6 \ A$.Substituting $I = 6 \ A$ into the power equation: $P_{\min} = \frac{0.2 	imes (6)^3}{6-4} = \frac{0.2 	imes 216}{2} = 0.2 	imes 108 = 21.6 \ W$.

The resistance of a device component decreases as the current through it increases and it is described by the relation,$R = \frac{0.2 I}{I-4}$,where $I$ is the current. Determine the minimum power delivered. (Assume $I > 4$) (in $W$)

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