91Ó°ÊÓ

$V_{\text{input}}=0.1\sin \mathrm{Ó\neg }t$

Resistance$=10\text{K}\mathrm{Î\copyright }$.

The highest current is $90\mathrm{μ}A$ and the lowest current is $70\mathrm{μ}A$.

The maximum power done by $V_{\mathrm{Î\pm }}$ is 216mW and the minimum power done by $V_{\mathrm{Î\pm }}$is168mW .

$$\begin{gathered} \mathbf{\text{Ohm'slaw:}}{\mathit{I}}_{\mathbf{h}}\mathbf{=}\frac{{\mathbf{V}}_{\stackrel{\mathbf{Ë™}}{\mathbf{h}}}}{\mathbf{R}} \\ \mathbf{\text{PowerdeliveredP=VI}} \\ \mathbf{\text{Powerdelivered}}\mathit{P}\mathbf{=}{\mathit{I}}^{\mathbf{2}}\mathit{R} \end{gathered}$$

(a)

The voltage across the resistor in the input circuit is expressed as,

.$V=V_{he}-0.7V-0.1V\sin \left(\mathrm{Ó\neg }t\right)$

Thus the highest voltage across the resistor would be:

$$\begin{gathered} V_h=V_{he}-0.7\text{V}-0.1\text{V}\sin \left(\mathrm{Ó\neg }t\right) \\ =1.5\text{V}-0.7\text{V}+0.1\text{V} \\ =0.9\text{V} \end{gathered}$$

$$\begin{gathered} V_I=V_{he}-0.7 \text{V}-0.1V\sin \left(\mathrm{Ó\neg }t\right) \\ =1.5\text{V}-0.7 \text{V}-0.1 \text{V} \\ =0.7V \end{gathered}$$

The resistor follows Ohm's Law, so the highest current is shown below.

$$\begin{gathered} I_h=\frac{V_h}{R} \\ =\frac{0.9V}{10K\mathrm{Î\copyright }} \\ =9×{10}^{-5} \text{A} \\ =90\mathrm{μ}A \end{gathered}$$

And the lowest current is given below.

$$\begin{gathered} I_I=\frac{V_I}{R} \\ =7×{10}^{-5} \text{A}\left[\frac{1\mathrm{μ}A}{{10}^{-6}A}\right] \\ \text{=7μ´¡} \end{gathered}$$

The highest current is $90\mathrm{μ}A$ and the lowest current is $70\mathrm{μ}A$.

(b)

The maximum power delivered by the input is shown below.

$$\begin{gathered} P_h=V{I}_h \\ =\left(0.1V\right)\left(90\mathrm{μ}A\right) \\ =9\mathrm{μ}A \end{gathered}$$

The minimum power delivered by the input is:

$$\begin{gathered} P_I=V{I}_I \\ =\left(0.1V\right)\left(70\mathrm{μ}A\right) \\ =7\mathrm{μ}A \end{gathered}$$

The maximum power delivered by the input is $\left|9\mathrm{μ}W\right|$ and the minimum power delivered by the P is$\left|7\mathrm{μ}W\right|$ .

(c)

The highest current in the output circuit is:

$$\begin{gathered} I_{n,h}=I_h×200 \\ =\left(90\mathrm{μ}A\right)\left(200\right) \\ =18mA \end{gathered}$$

So the lowest current in the output circuit is shown below.

$$\begin{gathered} I_{n,j}=I_h×200 \\ =\left(70\mathrm{μ}A\right)\left(200\right) \\ =14mA \end{gathered}$$

The maximum output power is shown below.

$$\begin{gathered} P=I_{n,h}^2{R}_n \\ =\left[18mA\left(\frac{{10}^{-3}A}{1mA}\right)\right]350\mathrm{Î\copyright } \\ =113mW \end{gathered}$$

The minimum output power is:

$$\begin{gathered} P=I_{n,J}^2{R}_n \\ =\left[14mA\left(\frac{{10}^{-3}A}{1mA}\right)\right]350\mathrm{Î\copyright } \\ =69mW \end{gathered}$$

The maximum output power is $113mW$ and the minimum output power is 69mW .

(d)

The maximum power done by $V_{\mathrm{Î\pm }}$ is given below.

role="math" localid="1658398499816" $$\begin{gathered} P_{\mathrm{Î\pm },h}=V_{\mathrm{Î\pm }}{I}_{\mathrm{Î\pm },h} \\ =\left(12v\right)\left(18mA\right) \\ =216mW \end{gathered}$$

The minimum power done by $V_{\mathrm{Î\pm }}$ is shown below.

$$\begin{gathered} P_{\mathrm{Î\pm },h}=V_{\mathrm{Î\pm }}{I}_{\mathrm{Î\pm },J} \\ =\left(12v\right)\left(14mA\right) \\ =168mW \end{gathered}$$

The maximum power done by$V_{\mathrm{Î\pm }}$is 216mW and the minimum power done by$V_a$is 168mW .

(e)

Power delivered$P=VI$.

Thus, greater the power greater will be work done.

The maximum power done by$V_{\mathrm{Î\pm }}$ is 216 mW and the minimum power done by$V_a$is 168mW .

Hence most work is done by the $V_a$.