The rate of change of current I (in mA)depends on time as the following: (dI)/(dt)=-(4)/(t)(t in s). Find the current as a function of time, if at the moment t=0.1 s the value of current I=12

To find the current as a function of time, we need to integrate the given rate of change of current with respect to time.<br /><br />Given:<br />$\frac{dI}{dt} = -\frac{4}{t}$<br /><br />Integrating both sides with respect to time:<br />$I = \int -\frac{4}{t} dt$<br /><br />$I = -4 \ln|t| + C$<br /><br />Now, we can use the initial condition to find the value of the constant $C$.<br /><br />At $t = 0.1$ s, $I = 12$ mA:<br />$12 = -4 \ln|0.1| + C$<br /><br />$C = 12 + 4 \ln|0.1|$<br /><br />Therefore, the current as a function of time is:<br />$I(t) = -4 \ln|t| + 12 + 4 \ln|0.1|$<br /><br />Simplifying the expression:<br />$I(t) = -4 \ln|t| + 12 + 4 \ln(0.1)$<br /><br />$I(t) = -4 \ln|t| + 12 - 4$<br /><br />$I(t) = -4 \ln|t| + 8$<br /><br />So, the current as a function of time is:<br />$I(t) = -4 \ln|t| + 8$