Answer:
[tex]\frac{dI}{dt} =-3.04*10^-^4amps/sec[/tex]
Explanation:
From the question we are told that
Voltage change [tex]\frac{dv}{dt}=-0.03volts/sec[/tex]
Resistance change [tex]\frac{dR}{dt}=0.02ohms/sec[/tex]
Resistance [tex]R=100[/tex]
Current [tex]I=0.02[/tex]
Generally the equation for ohms law is mathematically represented as
[tex]V=IR[/tex]
Therefore
[tex]\frac{dv}{dt} =R\frac{dI}{dt} +I\frac{dR}{dt}[/tex]
[tex]\frac{dv}{dt} =R\frac{dI}{dt} +I\frac{dR}{dt}[/tex]
Resolving the Rate of current changing [tex]\frac{dI}{dt}[/tex] as subject of formula
[tex]\frac{dv}{dt} =R\frac{dI}{dt} +I\frac{dR}{dt}[/tex]
[tex]R\frac{dI}{dt} =\frac{dv}{dt} -I\frac{dR}{dt}[/tex]
[tex]\frac{dI}{dt} =\frac{1}{R} (\frac{dv}{dt} -I\frac{dR}{dt})[/tex]
Therefore
[tex]\frac{dI}{dt} =\frac{(-0.03) -(0.02)(0.02)}{100}[/tex]
[tex]\frac{dI}{dt} =-3.04*10^-^4amps/sec[/tex]
Therefore the current decreases at a rate
[tex]\frac{dI}{dt} =3.04*10^-^4amps/sec[/tex]
