Calculate the energy in a circuit containing a resistor

Question

A voltage peak in a circuit is caused by a current through a resistor.

The energy E which is dissipated by the resistor is:

Calculate E if

Can anyone please give me some suggestions where to start ? What is the formula to use ?

Answer 1

Hint :
Use the Euler formula :
$$ sin(x)=\frac{exp(j\cdot x)-exp(-j\cdot x)}{2j}\\ $$ where $j^2=-1$

EDIT : $$ E = \int_{0}^{\infty} R I_{0}^{2}\cdot e^{-2t/t_{0}}\cdot sin^{2}(2t/t_{0}) \ dt $$ $$ E = \int_{0}^{\infty} R I_{0}^{2} e^{-2t/t_{0}}\cdot \frac{-2+exp(j4t/t_{0})+exp(-j4t/t_{0})}{-4} \ dt $$ $$ E = \frac{R I_{0}^{2}t_{0}}{-4}\cdot (-1 + \frac{1}{2-4j} + \frac{1}{2+4j}) $$ $$ E = \frac{R I_{0}^{2}t_{0}}{5} $$

By the way, downvoting without saying why is childish.

Answer 2

Since $R$ is a constant then the energy is seen to be \begin{align} E = R \int_{0}^{\infty} I^{2}(t) \ dt = I_{0}^{2} R \int_{0}^{\infty} e^{-2t/t_{0}} \ \sin^{2}(2t/t_{0}) \ dt \end{align} By making the substitution $x = 2t/t_{0}$ the integral becomes \begin{align} E &= \frac{I_{0}^{2} t_{0} R}{2} \int_{0}^{\infty} e^{-x} \ \sin^{2}(x) \ dx \\ &= \frac{I_{0}^{2} t_{0} R}{4} \int_{0}^{\infty} e^{-x} (1 - \cos(2x)) \ dx \\ &= \frac{I_{0}^{2} t_{0} R}{4} \left[ [- e^{-x} ]_{0}^{\infty} - \int_{0}^{\infty} e^{-x} \ \cos(2x) \ dx \right] \\ &= \frac{I_{0}^{2} t_{0} R}{4} \left[ 1 - \int_{0}^{\infty} e^{-x} \ \cos(2x) \ dx \right] \\ &= \frac{I_{0}^{2} t_{0} R}{4} \left( 1 - \frac{1}{5} \right) \\ E &= \frac{I_{0}^{2} t_{0} R}{5} \end{align}