Chapter 3, Problem 27E

Compute the power absorbed by each element of the circuit of Fig. 3.67.

■ FIGURE 3.67

To determine

Find power absorbed by each element in the circuit.

Answer to Problem 27E

Power absorbed by independent voltage source is $-1.818\text{mW}$, power absorbed by $1000\Omega$ resistor is $826.5\mu \text{W}$, power absorbed by dependent voltage source is $-1.240\text{mW}$, power absorbed by $2200\Omega$ resistor is $1.818\text{mW}$ and power absorbed by $500\Omega$ resistor is $413.2\mu \text{W}$.

Explanation of Solution

Formula used:

The expression for power absorbed by voltage source is as follows,

$p=vi$ (1)

Here,

$p$ is the power absorbed by voltage source,

$i$ is the current, and

$v$ is the voltage.

The expression for power absorbed by resistor is as follows,

$p=i^{2}R$ (2)

Here,

$p$ is the power absorbed,

$i$ is the current and

$R$ is value of resistance.

Calculation:

The circuit diagram is redrawn as shown in Figure 1,

Refer to the redrawn Figure 1,

The expression for KVL in mesh $DABCD$ is as follows,

$-v_1+i{R}_2+v_2+i{R}_3+i{R}_1=0$ (3)

Here,

$i$ is the current in the circuit,

$v_1$ and $v_2$ are the voltages in the circuit and

$R_1$, $R_2$ and $R_3$ are the resistances in the circuit.

The expression for voltage $v_x$ is as follows,

$v_x=-i{R}_1$ (4)

Here,

$v_x$ is the voltage,

$i$ is the current and

$R_1$ is value of resistance.

The expression for voltage $v_2$ is as follows,

$v_2=3v_x$ (5)

Here,

$v_x$ and $v_2$ are the voltages.

Refer to the redrawn Figure 1,

Substitute $2\text{V}$ for $v_1$, $3v_x$ for $v_2$, $500\text{Ω}$ for $R_1$, $1000\text{Ω}$ for $R_2$ and $2200\text{Ω}$ for $R_3$ in equation (3),

$-2\text{V}+i\left(1000\text{Ω}\right)+3v_x+i\left(2200\text{Ω}\right)+i\left(500\text{Ω}\right)=0$ (6)

Substitute $500\text{Ω}$ for $R_1$ in equation (4),

$v_x=-i\left(500\text{Ω}\right)$ (7)

Rearrange equation (6) and (7),

$\begin{array}{c}3700i+3v_x=2\\ 500i+v_x=0\end{array}$

The equations so formed can be written in matrix form as,

$\left(\begin{array}{cc}3700& 3\\ 500& 1\end{array}\right)\left(\begin{array}{c}i\\ v_x\end{array}\right)=\left(\begin{array}{c}2\\ 0\end{array}\right)$

Therefore, by Cramer’s rule,

The determinant of coefficient matrix is as follows,

$\begin{array}{c}\Delta =\left|\begin{array}{cc}3700& 3\\ 500& 1\end{array}\right|\\ =2200\end{array}$

The 1^st determinant is as follows,

$\begin{array}{c}{\Delta }_1=\left|\begin{array}{cc}2& 3\\ 0& 1\end{array}\right|\\ =2\end{array}$

The 2^nd determinant is as follows,

$\begin{array}{c}{\Delta }_2=\left|\begin{array}{cc}3700& 2\\ 500& 0\end{array}\right|\\ =-1000\end{array}$

Simplify for $i$,

$\begin{array}{c}i=\frac{{\Delta }_1}{\Delta }\\ =\frac{2}{2200}\text{ A}\\ =9.09\times {10}^{-4}\text{A}\end{array}$

Simplify for $v_x$,

$\begin{array}{c}{v}_x=\frac{{\Delta }_2}{\Delta }\\ =\frac{-1000}{2200}\text{ V}\\ =-0.454\text{V}\end{array}$

Substitute $-0.454\text{V}$ for $v_x$ in equation (5),

$\begin{array}{c}{v}_2=3\left(-0.454\text{V}\right)\\ =-1.364\text{V}\end{array}$

Current is leaving the positive terminal and we are calculating power absorbed means current should leave by negative terminal so we will use magnitude of voltage with negative sign, Thus the value of $v_1$ is $-2\text{V}$.

Substitute $-2\text{V}$ for $v_1$ and $9.09\times {10}^{-4}\text{A}$ for $i$ in equation (1),

$\begin{array}{c}p=\left(-2\text{V}\right)\left(9.09\times {10}^{-4}\text{A}\right)\\ =-1.818\times {10}^{-3}\text{W}\\ \text{=}-1.818\text{mW}\left\{\because {10}^{-3}\text{W}=1\text{mW}\right\}\end{array}$

So power absorbed by independent voltage source $v_1$ is $-1.818\text{mW}$.

Substitute $1000\text{Ω}$ for $R$ and $9.09\times {10}^{-4}\text{A}$ for $i$ in equation (2),

$\begin{array}{c}p={\left(9.09\times {10}^{-4}\text{A}\right)}^2\left(1000\text{Ω}\right)\\ =\left(82.65\times {10}^{-8}{\text{A}}^2\right)\left(1000\text{Ω}\right)\\ =826.5\times {10}^{-6}\text{W}\\ =826.5\mu \text{W}\left\{\because {10}^{-6}\text{W}=1\mu \text{W}\right\}\end{array}$

So power absorbed by resistor $R_2$ is $826.5\text{μW}$.

Substitute $-1.364\text{V}$ for $v_2$ and $9.09\times {10}^{-4}\text{A}$ for $i$ in equation (1),

$\begin{array}{c}p=\left(-1.364\text{V}\right)\left(9.09\times {10}^{-4}\text{A}\right)\\ =-1.240\times {10}^{-3}\text{W}\\ \text{=}-1.240\text{mW}\left\{\because {10}^{-3}\text{W}=1\text{mW}\right\}\end{array}$

So power absorbed by dependent voltage source $v_2$ is $-1.240\text{mW}$.

Substitute $2200\text{Ω}$ for $R$ and $9.09\times {10}^{-4}\text{A}$ for $i$ in equation (2),

$\begin{array}{c}p={\left(9.09\times {10}^{-4}\text{A}\right)}^2\left(2200\text{Ω}\right)\\ =\left(82.65\times {10}^{-8}{\text{A}}^2\right)\left(2200\text{Ω}\right)\\ =1.818\times {10}^{-3}\text{W}\\ =1.818\text{mW}\left\{\because {10}^{-3}\text{W}=1\text{mW}\right\}\end{array}$

So power absorbed by resistor $R_3$ is $1.818\text{mW}$.

Substitute $500\text{Ω}$ for $R$ and $9.09\times {10}^{-4}\text{A}$ for $i$ in equation (2),

$\begin{array}{c}p={\left(9.09\times {10}^{-4}\text{A}\right)}^2\left(500\text{Ω}\right)\\ =\left(82.65\times {10}^{-8}{\text{A}}^2\right)\left(500\text{Ω}\right)\\ =413.2\times {10}^{-6}\text{W}\\ =413.2\mu \text{W}\left\{\because {10}^{-6}\text{W}=1\mu \text{W}\right\}\end{array}$

So power absorbed by resistor $R_1$ is $413.2\mu \text{W}$.

Conclusion:

Thus, power absorbed by independent voltage source is $-1.818\text{mW}$, power absorbed by $1000\Omega$ resistor is $826.5\mu \text{W}$, power absorbed by dependent voltage source is $-1.240\text{mW}$, power absorbed by $2200\Omega$ resistor is $1.818\text{mW}$ and power absorbed by $500\Omega$ resistor is $413.2\mu \text{W}$.