Chapter 2, Problem 55E

Sketch the power absorbed by a 100 Ω resistor as a function of voltage over the range −2 V ≤ V_resistor ≤ + 2 V.

To determine

Sketch the power absorbed by $100\Omega$ resistor.

Explanation of Solution

Given data:

The value of resistor is $100\Omega$

The range of voltage is $\left(-2\text{ V}\le V_{resistor}\le +2\text{ V}\right)$

Formula used:

The expression for power across resistor is as follows.

$p_1=\left(\frac{v_1{}^2}{R}\right)$ (1)

Here,

$p_1$ is the power across resistor,

$v_1$ is the voltage between range $\left(-2\text{ V}\le V_{resistor}\le +2\text{ V}\right)$.

$R$ is the resistance.

The expression for power across resistor is as follows.

$p_2=\left(\frac{v_2{}^2}{R}\right)$ (2)

Here,

$p_2$ is the power across resistor,

$v_2$ is the voltage between range $\left(-2\text{ V}\le V_{resistor}\le +2\text{ V}\right)$.

The expression for power across resistor is as follows.

$p_3=\left(\frac{v_3{}^2}{R}\right)$ (3)

Here,

$p_3$ is the power across resistor,

$v_3$ is the voltage between range $\left(-2\text{ V}\le V_{resistor}\le +2\text{ V}\right)$.

The expression for power across resistor is as follows.

$p_4=\left(\frac{v_4{}^2}{R}\right)$ (4)

Here,

$p_4$ is the power across resistor.

$v_4$ is the voltage between range $\left(-2\text{ V}\le V_{resistor}\le +2\text{ V}\right)$.

The expression for power across resistor is as follows.

$p_5=\left(\frac{v_5{}^2}{R}\right)$ (5)

Here,

$p_5$ is the power across resistor.

$v_5$ is the voltage between range $\left(-2\text{ V}\le V_{resistor}\le +2\text{ V}\right)$.

The expression for power across resistor is as follows.

$p_6=\left(\frac{v_6{}^2}{R}\right)$ (6)

Here,

$p_6$ is the power across resistor,

$v_6$ is the voltage between range $\left(-2\text{ V}\le V_{resistor}\le +2\text{ V}\right)$.

The expression for power across resistor is as follows.

$p_7=\left(\frac{v_7{}^2}{R}\right)$ (7)

Here,

$p_7$ is the power across resistor,

$v_7$ is the voltage between range $\left(-2\text{ V}\le V_{resistor}\le +2\text{ V}\right)$.

The expression for power across resistor is as follows.

$p_8=\left(\frac{v_8{}^2}{R}\right)$ (8)

Here,

$p_8$ is the power across resistor,

$v_8$ is the voltage between range $\left(-2\text{ V}\le V_{resistor}\le +2\text{ V}\right)$.

Calculation:

Substitute $-2\text{ V}$ for $v_1$ and $100\Omega$ for $R$ in equation (1).

$\begin{array}{c}{p}_1=\left(\frac{{(-2\text{ V})}^2}{100\Omega }\right)\\ =0.04\text{ W}\\ \text{=40 mW }\left\{\because 1{\text{ W=10}}^3\text{ mW}\right\}\end{array}$

Substitute $-1\text{ V}$ for $v_2$ and $100\Omega$ for $R$ in equation (2).

$\begin{array}{c}{p}_2=\left(\frac{{(-1\text{ V})}^2}{100\Omega }\right)\\ =0.01\text{ W}\\ \text{=10 mW }\left\{\because 1{\text{ W=10}}^3\text{ mW}\right\}\end{array}$

Substitute $-0.5\text{ V}$ for $v_3$ and $100\Omega$ for $R$ in equation (3).

$\begin{array}{c}{p}_3=\left(\frac{{(-0.5\text{ V})}^2}{100\Omega }\right)\\ =2.5\times {10}^{-3}\text{ W}\\ \text{=2}\text{.5 mW }\left\{\because 1{\text{ W=10}}^3\text{ mW}\right\}\end{array}$

Substitute $\text{0 V}$ for $v_4$ and $100\Omega$ for $R$ in equation (4).

$\begin{array}{c}{p}_4=\left(\frac{{(0\text{ V})}^2}{100\Omega }\right)\\ =0\text{ W}\\ \text{=0 mW }\left\{\because 1{\text{ W=10}}^3\text{ mW}\right\}\end{array}$

Substitute $\text{1 V}$ for $v_5$ and $100\Omega$ for $R$ in equation (5).

$\begin{array}{c}{p}_5=\left(\frac{{(1\text{ V})}^2}{100\Omega }\right)\\ =0.01\text{ W}\\ \text{=10 mW }\left\{\because 1{\text{ W=10}}^3\text{ mW}\right\}\end{array}$

Substitute $0.5\text{ V}$ for $v_3$ and $100\Omega$ for $R$ in equation (3).

$\begin{array}{c}{p}_6=\left(\frac{{(0.5\text{ V})}^2}{100\Omega }\right)\\ =2.5\times {10}^{-3}\text{ W}\\ \text{=2}\text{.5 mW }\left\{\because 1{\text{ W=10}}^3\text{ mW}\right\}\end{array}$

Substitute $1\text{ V}$ for $v_2$ and $100\Omega$ for $R$ in equation (2).

$\begin{array}{c}{p}_7=\left(\frac{{(1\text{ V})}^2}{100\Omega }\right)\\ =0.01\text{ W}\\ \text{=10 mW }\left\{\because 1{\text{ W=10}}^3\text{ mW}\right\}\end{array}$

Substitute $2\text{ V}$ for $v_1$ and $100\Omega$ for $R$ in equation (1).

$\begin{array}{c}{p}_8=\left(\frac{{(2\text{ V})}^2}{100\Omega }\right)\\ =0.04\text{ W}\\ \text{=40 mW }\left\{\because 1{\text{ W=10}}^3\text{ mW}\right\}\end{array}$

The plot for the power absorbed by $100\Omega$-versus-voltage resistor is drawn in Figure 1.

Conclusion:

Thus, the sketch of the power absorbed by $100\Omega$ resistor is drawn in Figure 1.