Question

Assume that, as the battery wears out, the voltage decreases at 0.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.02 ohms per second. When the resistance is 100 ohms and the current is 0.02 amperes, at what rate is the current changing

Answer 1

Answer:

$\frac{dI}{dt} =-3*10^-^4amps/sec$

Explanation:

From the question we  are told that

Voltage decreases at   $\frac{dv}{dt} =-0.03volts/sec$

Resistance increase at $\frac{dR}{dt}=0.02ohms /sec$

Resistance at $R=100ohms$

Current at $I=0.02amps$

Generally the equation for ohms law is mathematically represented as

       $V=IR$

Therefore

       $\frac{dV}{dt} =R\frac{dI}{dt} +I\frac{dR}{dt}$

Generally making $\frac{dI}{dt}$ subject of the formula in the above equation mathematically gives

       $\frac{dV}{dt} =R\frac{dI}{dt} +I\frac{dR}{dt}$

      $R\frac{dI}{dt} = \frac{dV}{dt} -I\frac{dR}{dt}$

       $\frac{dI}{dt} =\frac{1}{R} (\frac{dV}{dt} -I\frac{dR}{dt})$

Therefore

       $\frac{dI}{dt} =\frac{1}{100}((-0.03) -(0.02)*(0.02))$

Generally it is given that the change in current is

       $\frac{dI}{dt} =-3*10^-^4amps/sec$