Question

The voltage and current at the terminals of the circuit element in Fig. \(1.5\) are zero for \(t<0\). For \(t \geq\) 0 they are

$$ \begin{array}{l} v=50 e^{-1600 t}-50 e^{-400 t} \mathrm{~V} \\ i=5 e^{-1600 t}-5 e^{-400 t} \mathrm{~mA} . \end{array} $$

a) Find the power at \(t=625 \mu \mathrm{s}\).

b) How much energy is delivered to the circuit element between 0 and 625 \mus.?

c) Find the total energy delivered to the element.

Answer 1

From the voltage and current equations, the power at any instant is given by

\(p(t)=v(t) i(t)=\left(50 e^{-1600 t}-50 e^{-400 t}\right)\left(5 e^{-1600 t}-5 e^{-400 t}\right)\)

\(\Rightarrow p(t)=250 e^{-3200 t}-500 e^{-2000 t}+250 e^{-800 t}\)

\(\Rightarrow p(t)=250\left[e^{-3200 t}-2 e^{-2000 t}+e^{-800 t}\right] m W\)

(b)

Now the energy delivered by the element for any time duration between \(t_{1}\) and \(t_{2}\) is given by

\(E=\int_{t_{1}}^{t_{2}} p(t) d t\)

in our case we need to find the energy delivered between 0 and 625 mu-s

So,

\(E=\int_{0}^{625 \mu s} 250\left[e^{-3200 t}-2 e^{-2000 t}+e^{-800 t}\right] d t\)

\(\Rightarrow E=250\left[\frac{e^{-3200 t}}{-3200}-2 \frac{e^{-2000 t}}{-2000}+\frac{e^{-800 t}}{-800}\right]_{0}^{625 \mu s}\)

\(\Rightarrow E=250\left[\left(\frac{e^{-2}}{-3200}-2 \frac{e^{-1.25}}{-2000}+\frac{e^{-0.5}}{-800}\right)-\left(\frac{1}{-3200}-2 \frac{1}{-2000}+\frac{1}{-800}\right)\right]\)

\(\Rightarrow E=250\left[\left(-5.1395 \times 10^{-4}\right)-\left(-5.625 \times 10^{-4}\right)\right]\)

\(\Rightarrow E=250\left[4.8549 \times 10^{-5}\right]\)

\(\Rightarrow E=0.0121 \mathrm{~mJ}\)

(c)

In order to find the total energy delivered by the element, we need to find the energy delivered for the duration of 0 sec to infinity.

\(E=\int_{0}^{\infty} 250\left[e^{-3200 t}-2 e^{-2000 t}+e^{-800 t}\right] d t\)

\(\Rightarrow E=250\left[\frac{e^{-3200 t}}{-3200}-2 \frac{e^{-2000 t}}{-2000}+\frac{e^{-800 t}}{-800}\right]_{0}^{\infty}\)

\(\Rightarrow E=250\left[\left(\frac{e^{-\infty}}{-3200}-2 \frac{e^{-\infty}}{-2000}+\frac{e^{-\infty}}{-800}\right)-\left(\frac{1}{-3200}-2 \frac{1}{-2000}+\frac{1}{-800}\right)\right]\)

\(\Rightarrow E=250\left[(0)-\left(-5.625 \times 10^{-4}\right)\right]\)

\(\Rightarrow E=250\left[5.625 \times 10^{-4}\right]\)

\(\Rightarrow E=0.3118 m J\)