Question

What power is dissipated by the $2\Omega$ resistor in $figureCP28.78$?

Short answer

Need to find the power is dissipated by the$2\Omega$.

Step-by-step solution

Given information

Need to find the power is dissipated by the$2\Omega$.

Simplify.

The frequency of the oscillation between $0Vand5V$is$f=10MHz$. The frequency is the reciprocal of the time period, So we get the period of the oscillation

$T=\frac{1}{f}$

Let us plug the value of $f$into equation (1) to get the period $T$

$T=\frac{1}{f}=\frac{1}{10\times {10}^{6}Hz}=0.1\times {10}^{-6}s$

At the high state $5V$, The oscillation spend $75\%$of the time ,So the period at the high state is

$T_H=0.75(0.1\times {10}^{-6}s)=75\times {10}^{-9}s$

The remaining$25\%$is spending at low state 0V. So, the period at low state is
localid="1648848680554" $T_L=0.25(0.1\times {10}^{-6}s=25\times {10}^{-9}s$

The period at the low is given in terms of $R_{2}andc$

$T_L=R_{2}C\times In2$

Our target is to find $R_2$, So we solve equation (2) for $R2$and plug the value for $Cand{T}_l$

$$\begin{gathered} R_2=\frac{T_l}{C\times In2} \\ =\frac{25\times {10}^{-9}s}{(500\times {10}^{-12}F)In2} \\ =72\Omega \end{gathered}$$ .

Calculation

The period at the high state is given in terms of $R_{2,}{R}_{1}andC$

$T_H=(R_1+R_2)C\times I{n}_2$

Our target is to find $R_1$, So we solve equation(3) for localid="1648850227432" $R_1$and plug the values for $C$, $R_{2}and{T}_H$

localid="1649106723568" $$\begin{gathered} R_1=\frac{T_H}{C\times I{n}_2}-R_2 \\ =\frac{75\times {10}^{-9}s}{(500\times {10}^{-12}F)I{n}_2}-72\Omega \\ =\boxed{144\Omega } \end{gathered}$$