Question

The voltage across a capacitor of 0.5F is defined by $V(t)=\begin{cases} 0, & t<0 \\ 2t, & 0<t<2s \\ 4e^{-(t-2)}, & t>2s \end{cases}$. Find $i(t)$

• A. \( -2e^{-(t-2)} \)
• B. \( -4e^{-(t-2)} \)
• C. \( -20e^{-(t-2)} \)
• D. \( -12e^{-(t-2)} \)

Hint:

The fundamental equation for current through a capacitor is $i(t) = C \frac{dV(t)}{dt}$. When dealing with piecewise functions for voltage, remember to differentiate each segment separately. Pay attention to the chain rule when differentiating exponential functions like $e^{-(t-2)}$. The derivative of $e^{f(t)}$ is $f'(t)e^{f(t)}$. In this case, $f(t) = -(t-2) = -t+2$, so $f'(t) = -1$.

Answer 1

The Correct Option is A

To find the current through the capacitor, we'll use the capacitor's voltage-current relationship, which states that the current is proportional to the rate of change of the voltage.

1. Understanding the Concepts:

- Capacitor Voltage-Current Relationship: The current through a capacitor is given by \( i(t) = C \frac{dV(t)}{dt} \), where \( C \) is the capacitance and \( V(t) \) is the voltage.
- Capacitance (C): The property of a capacitor to store electrical energy, measured in Farads (F).
- Voltage (V(t)): The potential difference across the capacitor as a function of time.
- Current (i(t)): The flow of charge onto the capacitor plates as a function of time.

2. Given Values:

\( C = 0.5 \text{ F} \)
\( V(t)=\begin{cases} 0, & t<0 \\ 2t, & 02s \end{cases} \)

3. Calculating the Current for t > 2s:

For \( t > 2s \), \( V(t) = 4e^{-(t-2)} \). Therefore, we need to find the derivative of \( V(t) \) with respect to \( t \):

\( \frac{dV(t)}{dt} = \frac{d}{dt} (4e^{-(t-2)}) = 4 \cdot \frac{d}{dt} (e^{-(t-2)}) = 4 \cdot (-1) e^{-(t-2)} = -4e^{-(t-2)} \)

Now, we can find \( i(t) \) using \( i(t) = C \frac{dV(t)}{dt} \):

\( i(t) = 0.5 \cdot (-4e^{-(t-2)}) = -2e^{-(t-2)} \)

Final Answer:

The current \( i(t) \) for \( t > 2s \) is \( -2e^{-(t-2)} \).