The electric current discharging in a R-C circuit is given by $i=i_0e^{\frac{-t}{RC}}$, where are $i_0, R, C$ are constant parameters.

Question

t is time. Let $i_0=2.00 units$, $R=(6.00)10^5 unit$ and $C=(0.500)10^{-6}$. Find the rate of change of current at t=0.3s

The answer need to be obtained by differentiating wrt t

$$\frac{di}{dt}=[i_0][e^{\frac{-t}{RC}}][\frac{-1}{RC}$$

Inputting the values $$-[2][e^{\frac{-0.3}{0.3}}][\frac{1}{0.3}]$$ Since $RC=(30)10^{-2}$

$$\frac{2}{0.3e}$$ I don’t think it can be simplified further. The right answer is, however, $\frac{2}{3e}$. Either I am wrong or the solution is wrong. Who is right?

(Derivative calculators don’t give a meaningful answer, please dont send me links, I have already tried them.)

Answer 1

I don't see any mistakes in your calculations (apart from not including a negative sign at the end as discussed below). Based on

$$i = i_0 e^{-t/(RC)} \tag{1}\label{eq1}$$

then with $i_0$ and $RC$ being constant, differentiating gives what you stated, i.e.,

$$\frac{di}{dt} = i_0 e^{-t/(RC)}\left(\frac{-1}{RC}\right) \tag{2}\label{eq2}$$

Thus, based on your provided values, $i_0 = 2$, $R = 6 \times 10^5$, $C = 0.5 \times 10^{-6}$ and $t = 0.3$, we get $RC = (6 \times 10^5) \times (0.5 \times 10^{-6}) = 3 \times 10^{-1} = 0.3$. Thus, $e^{-t/(RC)} = e^{-0.3/0.3} = e^{-1} = \frac{1}{e}$ and $\frac{-1}{RC} = \frac{-1}{0.3}$.

As such, the answer you gave of $\frac{2}{0.3e}$ should be $\frac{-2}{0.3e}$ (I assume you are just focusing on the absolute value part, i.e., by how much the current is decreasing). The "right" answer of $\frac{2}{3e}$ is $\frac{1}{10}$ of your answer and is incorrect.