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

The relationship between current and voltage for a capacitor is given by: $i(t) = C \frac{dV(t)}{dt}$ We are given the capacitance $C = 0.5$ F. The voltage $V(t)$ is a piecewise function. We need to find $i(t)$ for each interval.
Case 1: $t < 0$ $V(t) = 0$ $\frac{dV(t)}{dt} = \frac{d}{dt}(0) = 0$ $i(t) = C \times 0 = 0$ 
Case 2: $0 < t < 2s$ $V(t) = 2t$ $\frac{dV(t)}{dt} = \frac{d}{dt}(2t) = 2$ $i(t) = C \times 2 = 0.5 \times 2 = 1$ A 
Case 3: $t > 2s$ $V(t) = 4e^{-(t-2)}$ $\frac{dV(t)}{dt} = \frac{d}{dt}(4e^{-(t-2)})$ 
Using the chain rule, let $u = -(t-2) = -t+2$. Then $\frac{du}{dt} = -1$. 
So, $\frac{d}{dt}(4e^{-(t-2)}) = 4 \frac{d}{du}(e^u) \frac{du}{dt} = 4e^u (-1) = -4e^{-(t-2)}$ 
Now, calculate $i(t)$ for this interval: $i(t) = C \times \frac{dV(t)}{dt}$ $i(t) = 0.5 \times (-4e^{-(t-2)})$ $i(t) = -2e^{-(t-2)}$ 
A Combining these results, the current $i(t)$ is: $i(t) = \begin{cases} 0, & t < 0 \\1, & 0 < t < 2s \\-2e^{-(t-2)}, & t > 2s \end{cases}$ 
Looking at the options, we are likely asked for the current specifically for $t > 2s$, as this is the form of the given options. For $t > 2s$, $i(t) = -2e^{-(t-2)}$.