Answer:
[tex]\frac{dI}{dt} =-3*10^-^4amps/sec[/tex]
Explanation:
From the question we are told that
Voltage decreases at [tex]\frac{dv}{dt} =-0.03volts/sec[/tex]
Resistance increase at [tex]\frac{dR}{dt}=0.02ohms /sec[/tex]
Resistance at [tex]R=100ohms[/tex]
Current at [tex]I=0.02amps[/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]
Generally making [tex]\frac{dI}{dt}[/tex] subject of the formula in the above equation mathematically gives
[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{1}{100}((-0.03) -(0.02)*(0.02))[/tex]
Generally it is given that the change in current is
[tex]\frac{dI}{dt} =-3*10^-^4amps/sec[/tex]
