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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. Determined the minimum power deliver. (Assume, $I>4$ )
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$21.6 \mathrm{~W}$
Given, resistance $R$ of a device decreases as
current, $I$ increases by the relation,
$R=\frac{0.2 I}{I-4}$
$\therefore$ Power of the electric device, $P=I^2 R$
$=I^2 \cdot \frac{0.2 I}{I-4}$
Power will be minimum, if
$\begin{aligned} \frac{d P}{d I} & =0 \\ \frac{d}{d I}\left(\frac{0.2 I^3}{I-4}\right) & =0 \\ \frac{(I-4) \times 0.2 \times 3 I^2-0.2 I^3}{(I-4)^2} & =0 \\ 0.6 I^3-2.4 I^2-0.2 I^3 & =0 \\ 0.4 I^3-2.4 I^2 & =0 \\ 0.4 I^2(I-6) & =0 \\ I-6 & =0 \\ I \quad & =6 \mathrm{~A}\end{aligned} \quad(\because I>4)$
From Eq. (i), minimum power
$P_{\min }=\frac{0.2 \times 6^3}{6-4}=\frac{43.2}{2}=21.6 \mathrm{~W}$
Hence, the minimum power delivered in a device $21.6 \mathrm{~W}$.
current, $I$ increases by the relation,
$R=\frac{0.2 I}{I-4}$
$\therefore$ Power of the electric device, $P=I^2 R$
$=I^2 \cdot \frac{0.2 I}{I-4}$
Power will be minimum, if
$\begin{aligned} \frac{d P}{d I} & =0 \\ \frac{d}{d I}\left(\frac{0.2 I^3}{I-4}\right) & =0 \\ \frac{(I-4) \times 0.2 \times 3 I^2-0.2 I^3}{(I-4)^2} & =0 \\ 0.6 I^3-2.4 I^2-0.2 I^3 & =0 \\ 0.4 I^3-2.4 I^2 & =0 \\ 0.4 I^2(I-6) & =0 \\ I-6 & =0 \\ I \quad & =6 \mathrm{~A}\end{aligned} \quad(\because I>4)$
From Eq. (i), minimum power
$P_{\min }=\frac{0.2 \times 6^3}{6-4}=\frac{43.2}{2}=21.6 \mathrm{~W}$
Hence, the minimum power delivered in a device $21.6 \mathrm{~W}$.
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