Question #28493

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Question #28493

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Answer 1 · 2017-02-24T14:01:26

(a) $- 1 . ¯ ¯¯ ¯ 81$ $-1.bar81$
(b) $- 2$ $-2$
(c) $- 0 . ¯ 8$ $-0.bar8$

Explanation:

(a) Given $V = I R$ $V=IR$
For $R = 1,$ $R=1,$ $I = \frac{2}{1} = 2 A$ $I=2/1=2A$
For $R = 1.1,$ $R=1.1,$ $I = \frac{2}{1.1} = 1 . ¯ ¯¯ ¯ 81 A$ $I=2/1.1=1.bar81A$
$:.$ the average rate of change of $I$ with respect to $R$ in the interval of interest is $=(2-1.bar81)/(1-1.1)=-1.bar81$

(b) $I=V/R$
$=>I=2/R$
To find rate of change we differentiate function $I$ with respect to Resistance $R$ .
$(dI)/(dR)=2xx(-1)R^-2$
$=>(dI)/(dR)=-2R^-2$
ROC at $R=1Omega$
$(dI)/(dR)|_1=-2xx(1)^-2$
$=>(dI)/(dR)=-2$

(c) $R=V/I$
$=>R=2/I$
To find rate of change we differentiate function $R$ with respect to Current $I$
$=>(dR)/(dI)=-2I^-2$
ROC at $I=1.5A$
$(dR)/(dI)|_1.5=-2xx(1.5)^-2$
$(dR)/(dI)=-0.bar8$

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Question #28493

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