Answer
${\dfrac{dV}{dt}}=1 {\dfrac{volt}{sec}}$;
${\dfrac{dI}{dt}}={\dfrac{-1}{3}} {\dfrac{ampere}{sec}}$;
${\dfrac{dR}{dt}}={\dfrac{1}{I}({\dfrac{dV}{dt}-{\dfrac{V}{I}{\dfrac{dI}{dt}}}})}$;
and
${\dfrac{dR}{dt}}={\dfrac{3}{2}{\dfrac{ohm}{sec}}}$
Work Step by Step
a) Rate of increase in voltage is 1 volt per sec, then
${\dfrac{dV}{dt}}=1 {\dfrac{volt}{sec}}$
b) Rate of decreasing in current is given by
${\frac{dI}{dt}}={\frac{-1}{3}} {\dfrac{ampere}{sec}}$
c) Since, $V=IR$
Then $R={\dfrac{V}{I}}$
Now,${\dfrac{dR}{dt}}={\frac{I\frac{dV}{dt}-{V\frac{dI}{dt}}}{I^2}}$
${\frac{dR}{dt}}={\frac{1}{I}({\frac{dV}{dt}-{\frac{V}{I}{\frac{dI}{dt}}}})}$
d) From above parts a, b, c, we get
${\dfrac{dR}{dt}}={\frac{1}{I}({\frac{dV}{dt}-{\frac{V}{I}{\frac{dI}{dt}}}})}={\frac{1}{2}{(1-{\frac{12}{2}{\frac{-1}{3}}})}}$
where $V=12$ volt and $I =2$ amp
So, ${\dfrac{dR}{dt}}={\dfrac{3}{2}{\dfrac{ohm}{sec}}}$
