Chapter 1

Q. 1.P.3

The current through 10 Ω resistor is shown in the following figure. Determine the power dissipated in the resistor.

1.1.3

Step-by-Step

Verified Solution

Average power is given by

\begin{aligned} P_{\text {avg }} & =\frac{\text { Energy absorbed in one period }}{\text { period }} \\ E_R & =\int i^2(t) \cdot R d t=\frac{1}{T} \int_0^T i^2(t) R d t=I_{\text {rms }}^2 \cdot R \end{aligned}

Therefore,

\begin{aligned} i(t)=\frac{(10-0)}{(1-0)} \cdot & t=10 t \quad 0 \leq t \leq 1 \quad 0 \leq t \leq 1 \\ E_R & =\int_0^1(10 t)^2 \cdot 10 d t \\ & =1000\left|\frac{t^3}{3}\right|_0^1 \\ & =1000\left\lgroup \frac{1}{3} \right\rgroup=\frac{1000}{3}  J \end{aligned}

Therefore,

P_{ avg }=\frac{\frac{1000}{3}}{1}=\frac{1000}{3}  W

Peak power (10)^2 \cdot 10=1000  W