The electric curren in a charging R-C circuit is given by `i=i_0e^(-t/RC)` when `i_0`, R and C aere constant parameters of the circuit and t is time. Find the rate of change of current at `a. t=0, b. t=RC c. t=10RC`.

Explanation

To find the rate of change of current in the given R-C circuit at different time intervals, we start with the expression for current:

\[ i(t) = i_0 e^{-t/(RC)} \]

where \( i_0 \) is the initial current, \( R \) is the resistance, \( C \) is the capacitance, and \( t \) is time.

### Step 1: Differentiate the current with respect to time

To find the rate of change of current, we need to differentiate \( i(t) \) with respect to \( t \):

\[ \frac{di}{dt} = \frac{d}{dt}(i_0 e^{-t/(RC)}) \]

Using the chain rule, we have:

\[ \frac{di}{dt} = i_0 \cdot \frac{d}{dt}(e^{-t/(RC)}) = i_0 \cdot e^{-t/(RC)} \cdot \left(-\frac{1}{RC}\right) \]

Thus, we can express the derivative as:

\[ \frac{di}{dt} = -\frac{i_0}{RC} e^{-t/(RC)} \]

### Step 2: Evaluate at \( t = 0 \)

Now, we will evaluate the rate of change of current at \( t = 0 \):

\[ \frac{di}{dt} \bigg|_{t=0} = -\frac{i_0}{RC} e^{-0/(RC)} = -\frac{i_0}{RC} \cdot 1 = -\frac{i_0}{RC} \]

### Step 3: Evaluate at \( t = RC \)

Next, we evaluate the rate of change of current at \( t = RC \):

\[ \frac{di}{dt} \bigg|_{t=RC} = -\frac{i_0}{RC} e^{-RC/(RC)} = -\frac{i_0}{RC} e^{-1} = -\frac{i_0}{RC e} \]

### Step 4: Evaluate at \( t = 10RC \)

Finally, we evaluate the rate of change of current at \( t = 10RC \):

\[ \frac{di}{dt} \bigg|_{t=10RC} = -\frac{i_0}{RC} e^{-10RC/(RC)} = -\frac{i_0}{RC} e^{-10} = -\frac{i_0}{RC e^{10}} \]

### Summary of Results

- At \( t = 0 \): \( \frac{di}{dt} = -\frac{i_0}{RC} \)
- At \( t = RC \): \( \frac{di}{dt} = -\frac{i_0}{RC e} \)
- At \( t = 10RC \): \( \frac{di}{dt} = -\frac{i_0}{RC e^{10}} \)