Chapter 11, Problem 3E

Calculate the power absorbed at t = 0⁻, t = 0+, and t = 200 ms by each of the elements in the circuit of Fig. 11.27 if v_s is equal to (a) −10u(−t) V; (b) 20 + 5u(t) V.

■ FIGURE 11.27

To determine

Find the power absorbed at $t=0^{-}$, $t=0^{+}$, and $t=200\text{ms}$ by each of the three elements given in the circuit in Figure 11.27 in the textbook.

Answer to Problem 3E

The power absorbed at $t=0^{-}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{-100\text{W}}$, $\underset{\_}{100\text{W}}$, and $\underset{\_}{0\text{W}}$, respectively.

The power absorbed at $t=0^{+}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{0\text{W}}$, $\underset{\_}{100\text{W}}$, and $\underset{\_}{-100\text{W}}$, respectively.

The power absorbed at $t=200\text{ms}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{0\text{W}}$, $\underset{\_}{20.19\text{W}}$, and $\underset{\_}{-20.19\text{W}}$, respectively.

Explanation of Solution

Given data:

Refer to Figure 11.27 in the textbook for the given circuit.

The circuit parameters are given as follows:

$\begin{array}{l}{v}_s\left(t\right)=-10u\left(-t\right)\text{V}\\ R=1\Omega \\ L=250\text{mH}\end{array}$

Formula used:

Write the expression for power absorbed by the voltage source as follows:

$p_s\left(t\right)=-v_s\left(t\right)i\left(t\right)$        (1)

Here,

$v_s\left(t\right)$ is the instantaneous source voltage and

$i\left(t\right)$ is the instantaneous current through the source (through each element as the circuit is series circuit).

Write the expression for power absorbed by the resistor in the given circuit as follows:

$p_R\left(t\right)={\left[i\left(t\right)\right]}^{2}R$        (2)

Here,

$R$ is the resistance of the resistor.

Write the expression for conservation power in the circuit as follows:

${\displaystyle \sum p\left(t\right)}=0$        (3)

Here,

$p\left(t\right)$ is the total absorbed in the given circuit.

Calculation:

Find the source voltage at $t=0^{-}$ as follows:

$\begin{array}{c}{v}_s\left(0^{-}\right)=-10u\left(-0^{-}\right)\text{V}\\ =-10\left(1\right)\text{V}\left\{\because u\left(-0^{-}\right)=1\right\}\\ =-10\text{V}\end{array}$

Write the expression for current through each element in the given circuit as follows:

$i\left(t\right)=I_0{e}^{-\frac{t}{\tau }}$        (4)

Here,

$I_0$ is the initial value of the current and

$\tau$ is the time constant.

Write the expression for time constant in the given circuit as follows:

$\tau =\frac{L}{R}$

Substitute $250\text{mH}$ for $L$ and $1\Omega$ for $R$ to obtain the time constant in the given circuit.

$\begin{array}{c}\tau =\frac{250\text{mH}}{1\Omega }\\ =\frac{250\times {10}^{-3}}{1}\text{s}\\ =0.25\text{s}\end{array}$

From the given circuit, the initial value of the current is determined as follows:

$I_0=\frac{V_0}{R}$

From the given data, substitute $-10\text{V}$ for $V_0$ and $1\Omega$ for $R$ to obtain the initial value of the current as follows:

$\begin{array}{c}{I}_0=\frac{-10\text{V}}{1\Omega }\\ =-10\text{A}\end{array}$

Substitute $-10\text{A}$ for $I_0$ and 0.25 s for $\tau$ in Equation (4) to obtain the current through each element in the given circuit.

$i\left(t\right)=\left(-10\text{A}\right)e^{-\frac{t}{0.25\text{s}}}$

$i\left(t\right)=-10e^{-4t}\text{A}$        (5)

Find the current $i\left(t\right)$ at $t=0^{-}$ as follows:

$\begin{array}{c}i\left(0^{-}\right)=-10e^{-4\left(0\right)}\text{A}\\ =-10\left(1\right)\text{A}\\ =-10\text{A}\end{array}$

Modify the expression in Equation (1) for $t=0^{-}$ as follows:

$p_s\left(0^{-}\right)=-v_s\left(0^{-}\right)i\left(0^{-}\right)$

Substitute $-10\text{V}$ for $v_s\left(0^{-}\right)$ and $-10\text{A}$ for $i\left(0^{-}\right)$ to obtain the power absorbed by the voltage source at $t=0^{-}$.

$\begin{array}{c}{p}_s\left(0^{-}\right)=-\left(-10\text{V}\right)\left(-10\text{A}\right)\\ =-100\text{W}\end{array}$

Modify the expression in Equation (2) for $t=0^{-}$ as follows:

$p_R\left(0^{-}\right)={\left[i\left(0^{-}\right)\right]}^{2}R$

Substitute $-10\text{A}$ for $i\left(0^{-}\right)$ and $1\Omega$ for $R$ to obtain the value of $p_R\left(0^{-}\right)$.

$\begin{array}{c}{p}_R\left(0^{-}\right)={\left(-10\text{A}\right)}^2\left(1\Omega \right)\\ =100\text{W}\end{array}$

Rewrite the expression in Equation (3) as follows:

$p_s\left(t\right)+p_R\left(t\right)+p_L\left(t\right)=0$        (6)

Here,

$p_L\left(t\right)$ is the power absorbed by the inductor.

Rewrite the expression in Equation (6) for the power absorbed by the inductor as follows:

$p_L\left(t\right)=-\left[p_s\left(t\right)+p_R\left(t\right)\right]$        (7)

Modify the expression for $t=0^{-}$ as follows:

$p_L\left(0^{-}\right)=-\left[p_s\left(0^{-}\right)+p_R\left(0^{-}\right)\right]$

Substitute $-100\text{W}$ for $p_s\left(0^{-}\right)$ and 100 W for $p_R\left(0^{-}\right)$ to obtain the value of power absorbed by the inductor at $t=0^{-}$.

$\begin{array}{c}{p}_L\left(0^{-}\right)=-\left[-100\text{W}+100\text{W}\right]\\ =0\text{W}\end{array}$

Find the source voltage at $t=0^{+}$ as follows:

$\begin{array}{c}{v}_s\left(0^{+}\right)=-10u\left(-0^{+}\right)\text{V}\\ =-10\left(0\right)\text{V}\left\{\because u\left(-0^{+}\right)=0\right\}\\ =0\text{V}\end{array}$

Substitute $0^{+}$ for $t$ in Equation (4) to find the current $i\left(t\right)$ at $t=0^{+}$ as follows:

$\begin{array}{c}i\left(0^{+}\right)=-10e^{-4\left(0\right)}\text{A}\\ =-10\left(1\right)\text{A}\\ =-10\text{A}\end{array}$

Modify the expression in Equation (1) for $t=0^{+}$ as follows:

$p_s\left(0^{+}\right)=-v_s\left(0^{+}\right)i\left(0^{+}\right)$

Substitute 0 V for $v_s\left(0^{+}\right)$ and $-10\text{A}$ for $i\left(0^{+}\right)$ to obtain the power absorbed by the voltage source at $t=0^{+}$.

$\begin{array}{c}{p}_s\left(0^{+}\right)=-\left(0\text{V}\right)\left(-10\text{A}\right)\\ =0\text{W}\end{array}$

Modify the expression in Equation (2) for $t=0^{+}$ as follows:

$p_R\left(0^{+}\right)={\left[i\left(0^{+}\right)\right]}^{2}R$

Substitute $-10\text{A}$ for $i\left(0^{+}\right)$ and $1\Omega$ for $R$ to obtain the value of $p_R\left(0^{+}\right)$.

$\begin{array}{c}{p}_R\left(0^{+}\right)={\left(-10\text{A}\right)}^2\left(1\Omega \right)\\ =100\text{W}\end{array}$

Modify the expression in Equation (7) for $t=0^{+}$ as follows:

$p_L\left(0^{+}\right)=-\left[p_s\left(0^{+}\right)+p_R\left(0^{+}\right)\right]$

Substitute 0 W for $p_s\left(0^{+}\right)$ and 100 W for $p_R\left(0^{+}\right)$ to obtain the value of power absorbed by the inductor at $t=0^{+}$.

$\begin{array}{c}{p}_L\left(0^{+}\right)=-\left[0\text{W}+100\text{W}\right]\\ =-100\text{W}\end{array}$

Find the source voltage at $t=200\text{ms}$ as follows:

$\begin{array}{c}{v}_s\left(200\text{ms}\right)=-10u\left(-200\text{ms}\right)\text{V}\\ =-10\left(0\right)\text{V}\left\{\because u\left(-200\text{ms}\right)=0\right\}\\ =0\text{V}\end{array}$

Substitute $200\text{ms}$ for $t$ in Equation (4) to find the current $i\left(t\right)$ at $t=200\text{ms}$ as follows:

$\begin{array}{c}i\left(200\text{ms}\right)=-10e^{-4\left(200\text{ms}\right)}\text{A}\\ =-10e^{-4\left(0.2\right)}\text{A}\\ =-4.4932\text{A}\end{array}$

Modify the expression in Equation (1) for $t=200\text{ms}$ as follows:

$p_s\left(200\text{ms}\right)=-v_s\left(200\text{ms}\right)i\left(200\text{ms}\right)$

Substitute 0 V for $v_s\left(200\text{ms}\right)$ and $-4.4932\text{A}$ for $i\left(200\text{ms}\right)$ to obtain the power absorbed by the voltage source at $t=200\text{ms}$.

$\begin{array}{c}{p}_s\left(200\text{ms}\right)=-\left(0\text{V}\right)\left(-4.4932\text{A}\right)\\ =0\text{W}\end{array}$

Modify the expression in Equation (2) for $t=200\text{ms}$ as follows:

$p_R\left(200\text{ms}\right)={\left[i\left(200\text{ms}\right)\right]}^{2}R$

Substitute $-4.4932\text{A}$ for $i\left(200\text{ms}\right)$ and $1\Omega$ for $R$ to obtain the value of $p_R\left(200\text{ms}\right)$.

$\begin{array}{c}{p}_R\left(200\text{ms}\right)={\left(-4.4932\text{A}\right)}^2\left(1\Omega \right)\\ =20.1888\text{W}\\ \cong 20.19\text{W}\end{array}$

Modify the expression in Equation (7) for $t=200\text{ms}$ as follows:

$p_L\left(200\text{ms}\right)=-\left[p_s\left(200\text{ms}\right)+p_R\left(200\text{ms}\right)\right]$

Substitute 0 W for $p_s\left(200\text{ms}\right)$ and 20.19 W for $p_R\left(200\text{ms}\right)$ to obtain the value of power absorbed by the inductor at $t=200\text{ms}$.

$\begin{array}{c}{p}_L\left(200\text{ms}\right)=-\left[0\text{W}+20.19\text{W}\right]\\ =-20.19\text{W}\end{array}$

Conclusion:

Thus, the power absorbed at $t=0^{-}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{-100\text{W}}$, $\underset{\_}{100\text{W}}$, and $\underset{\_}{0\text{W}}$, respectively.

The power absorbed at $t=0^{+}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{0\text{W}}$, $\underset{\_}{100\text{W}}$, and $\underset{\_}{-100\text{W}}$, respectively.

The power absorbed at $t=200\text{ms}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{0\text{W}}$, $\underset{\_}{20.19\text{W}}$, and $\underset{\_}{-20.19\text{W}}$, respectively.

To determine

Find the power absorbed at $t=0^{-}$, $t=0^{+}$, and $t=200\text{ms}$ by each of the three elements given in the circuit in Figure 11.27 in the textbook.

Answer to Problem 3E

The power absorbed at $t=0^{-}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{-400\text{W}}$, $\underset{\_}{400\text{W}}$, and $\underset{\_}{0\text{W}}$, respectively.

The power absorbed at $t=0^{+}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{-500\text{W}}$, $\underset{\_}{400\text{W}}$, and $\underset{\_}{100\text{W}}$, respectively.

The power absorbed at $t=200\text{ms}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{-568.83\text{W}}$, $\underset{\_}{517.71\text{W}}$, and $\underset{\_}{51.12\text{W}}$, respectively.

Explanation of Solution

Given data:

The source voltage is given as follows:

$v_s\left(t\right)=20+5u\left(t\right)\text{V}$

Calculation:

Find the source voltage at $t=0^{-}$ as follows:

$\begin{array}{c}{v}_s\left(0^{-}\right)=20+5u\left(0^{-}\right)\text{V}\\ =20+5\left(0\right)\text{V}\left\{\because u\left(0^{-}\right)=0\right\}\\ =20\text{V}\end{array}$

Write the expression for current through each element in the given circuit as follows:

$i\left(t\right)=20+5\left(1-e^{-\frac{t}{\tau }}\right)$

From Part (a), substitute 0.25 s for $\tau$ to obtain the current through each element in the given circuit.

$i\left(t\right)=20+5\left(1-e^{-\frac{t}{0.25\text{s}}}\right)$

$i\left(t\right)=20+5\left(1-e^{-4t}\right)\text{A}$        (8)

Find the current $i\left(t\right)$ at $t=0^{-}$ as follows:

$\begin{array}{c}i\left(0^{-}\right)=20+5\left(1-e^{-4\left(0\right)}\right)\text{A}\\ =20+5\left(0\right)\text{A}\\ =20\text{A}\end{array}$

Modify the expression in Equation (1) for $t=0^{-}$ as follows:

$p_s\left(0^{-}\right)=-v_s\left(0^{-}\right)i\left(0^{-}\right)$

Substitute 20 V for $v_s\left(0^{-}\right)$ and 20 A for $i\left(0^{-}\right)$ to obtain the power absorbed by the voltage source at $t=0^{-}$.

$\begin{array}{c}{p}_s\left(0^{-}\right)=-\left(20\text{V}\right)\left(20\text{A}\right)\\ =-400\text{W}\end{array}$

Modify the expression in Equation (2) for $t=0^{-}$ as follows:

$p_R\left(0^{-}\right)={\left[i\left(0^{-}\right)\right]}^{2}R$

Substitute 20 A for $i\left(0^{-}\right)$ and $1\Omega$ for $R$ to obtain the value of $p_R\left(0^{-}\right)$.

$\begin{array}{c}{p}_R\left(0^{-}\right)={\left(20\text{A}\right)}^2\left(1\Omega \right)\\ =400\text{W}\end{array}$

Substitute $-400\text{W}$ for $p_s\left(0^{-}\right)$ and 400 W for $p_R\left(0^{-}\right)$ in Equation (7) to obtain the value of power absorbed by the inductor at $t=0^{-}$.

$\begin{array}{c}{p}_L\left(0^{-}\right)=-\left[-400\text{W}+400\text{W}\right]\\ =0\text{W}\end{array}$

Find the source voltage at $t=0^{+}$ as follows:

$\begin{array}{c}{v}_s\left(0^{+}\right)=20+5u\left(0^{+}\right)\text{V}\\ =20+5\left(1\right)\text{V}\left\{\because u\left(0^{+}\right)=1\right\}\\ =25\text{V}\end{array}$

Substitute $0^{+}$ for $t$ in Equation (8) to find the current $i\left(t\right)$ at $t=0^{+}$ as follows:

$\begin{array}{c}i\left(0^{+}\right)=20+5\left(1-e^{-4\left(0\right)}\right)\text{A}\\ =20+5\left(0\right)\text{A}\\ =20\text{A}\end{array}$

Modify the expression in Equation (1) for $t=0^{+}$ as follows:

$p_s\left(0^{+}\right)=-v_s\left(0^{+}\right)i\left(0^{+}\right)$

Substitute 25 V for $v_s\left(0^{+}\right)$ and 20 for $i\left(0^{+}\right)$ to obtain the power absorbed by the voltage source at $t=0^{+}$.

$\begin{array}{c}{p}_s\left(0^{+}\right)=-\left(25\text{V}\right)\left(20\text{A}\right)\\ =-500\text{W}\end{array}$

Modify the expression in Equation (2) for $t=0^{+}$ as follows:

$p_R\left(0^{+}\right)={\left[i\left(0^{+}\right)\right]}^{2}R$

Substitute 20 A for $i\left(0^{+}\right)$ and $1\Omega$ for $R$ to obtain the value of $p_R\left(0^{+}\right)$.

$\begin{array}{c}{p}_R\left(0^{+}\right)={\left(20\text{A}\right)}^2\left(1\Omega \right)\\ =400\text{W}\end{array}$

Modify the expression in Equation (7) for $t=0^{+}$ as follows:

$p_L\left(0^{+}\right)=-\left[p_s\left(0^{+}\right)+p_R\left(0^{+}\right)\right]$

Substitute $-500\text{W}$ for $p_s\left(0^{+}\right)$ and 400 W for $p_R\left(0^{+}\right)$ to obtain the value of power absorbed by the inductor at $t=0^{+}$.

$\begin{array}{c}{p}_L\left(0^{+}\right)=-\left[-500\text{W}+400\text{W}\right]\\ =100\text{W}\end{array}$

Find the source voltage at $t=200\text{ms}$ as follows:

$\begin{array}{c}{v}_s\left(200\text{ms}\right)=20+5u\left(200\text{ms}\right)\text{V}\\ =20+5\left(1\right)\text{V}\left\{\because u\left(200\text{ms}\right)=1\right\}\\ =25\text{V}\end{array}$

Substitute $200\text{ms}$ for $t$ in Equation (8) to find the current $i\left(t\right)$ at $t=200\text{ms}$ as follows:

$\begin{array}{c}i\left(200\text{ms}\right)=20+5\left(1-e^{-4\left(200\text{ms}\right)}\right)\\ =20+5\left(1-e^{-0.8}\right)\text{A}\\ =22.7533\text{A}\end{array}$

Modify the expression in Equation (1) for $t=200\text{ms}$ as follows:

$p_s\left(200\text{ms}\right)=-v_s\left(200\text{ms}\right)i\left(200\text{ms}\right)$

Substitute 25 V for $v_s\left(200\text{ms}\right)$ and 22.7533 for $i\left(200\text{ms}\right)$ to obtain the power absorbed by the voltage source at $t=200\text{ms}$.

$\begin{array}{c}{p}_s\left(200\text{ms}\right)=-\left(25\text{V}\right)\left(22.7533\text{A}\right)\\ =-568.8325\text{W}\\ \cong -568.83\text{W}\end{array}$

Modify the expression in Equation (2) for $t=200\text{ms}$ as follows:

$p_R\left(200\text{ms}\right)={\left[i\left(200\text{ms}\right)\right]}^{2}R$

Substitute 22.7533 A for $i\left(200\text{ms}\right)$ and $1\Omega$ for $R$ to obtain the value of $p_R\left(200\text{ms}\right)$.

$\begin{array}{c}{p}_R\left(200\text{ms}\right)={\left(22.7533\text{A}\right)}^2\left(1\Omega \right)\\ =517.7126\text{W}\\ \cong 517.71\text{W}\end{array}$

Modify the expression in Equation (7) for $t=200\text{ms}$ as follows:

$p_L\left(200\text{ms}\right)=-\left[p_s\left(200\text{ms}\right)+p_R\left(200\text{ms}\right)\right]$

Substitute $-568.83\text{W}$ for $p_s\left(200\text{ms}\right)$ and 517.71 W for $p_R\left(200\text{ms}\right)$ to obtain the value of power absorbed by the inductor at $t=200\text{ms}$.

$\begin{array}{c}{p}_L\left(200\text{ms}\right)=-\left[-568.83\text{W}+517.71\text{W}\right]\\ =51.12\text{W}\end{array}$

Conclusion:

Thus, the power absorbed at $t=0^{-}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{-400\text{W}}$, $\underset{\_}{400\text{W}}$, and $\underset{\_}{0\text{W}}$, respectively.

The power absorbed at $t=0^{+}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{-500\text{W}}$, $\underset{\_}{400\text{W}}$, and $\underset{\_}{100\text{W}}$, respectively.

The power absorbed at $t=200\text{ms}$ by source voltage $v_s$, resistor, and inductor are $\underset{\_}{-568.83\text{W}}$, $\underset{\_}{517.71\text{W}}$, and $\underset{\_}{51.12\text{W}}$, respectively.