Chapter 3, Problem 60P

To determine

Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Answer to Problem 60P

The power dissipated in $1\Omega$, $2\Omega$, $4\Omega$, and $8\Omega$ resistors are $1600\text{W}$, $3872\text{W}$, $14400\text{W}$, and $4608\text{W}$ respectively.

Explanation of Solution

Given data:

Refer to Figure 3.104 in the textbook for the nodal analysis. In the given circuit, $i_o$ is the current flow through $1\Omega$ resistor.

Calculation:

Modify the given figure with the representation of node voltage as shown in Figure 1.

Apply Kirchhoff’s current law at node $v_1$ in Figure 1.

$\frac{v_1-0}{1}+\frac{v_1-280}{4}+0.5i_o=0$ (1)

With reference to Figure 1, write the expression for $i_o$ through $1\Omega$ branch.

$i_o=\frac{v_1-0}{1}$ (2)

Substitute equation (2) in (1).

$\begin{array}{l}\frac{v_1-0}{1}+\frac{v_1-280}{4}+0.5\left(\frac{v_1-0}{1}\right)=0\\ v_1+\frac{v_1}{4}-\frac{280}{4}+0.5v_1=0\\ v_1+0.25v_1+0.5v_1=\frac{280}{4}\\ v_1+0.25v_1+0.5v_1=70\end{array}$

Simplify the equation as follows.

$\begin{array}{l}{v}_1+0.25v_1+0.5v_1=70\\ 1.75v_1=70\\ v_1=\frac{70}{1.75}\\ v_1=40\text{V}\end{array}$

Apply Kirchhoff’s current law at node $v_2$ in Figure 1.

$\frac{v_2-280}{8}+\frac{v_2-0}{2}-0.5i_o=0$ (3)

Substitute equation (2) in (3).

$\begin{array}{l}\frac{v_2-280}{8}+\frac{v_2-0}{2}-0.5\left(\frac{v_1-0}{1}\right)=0\\ \frac{v_2}{8}-\frac{280}{8}+\frac{v_2}{2}-0.5v_1=0\\ -0.5v_1+\left(\frac{1}{8}+\frac{1}{2}\right)v_2=\frac{280}{8}\end{array}$

$-0.5v_1+0.625v_2=35$ (4)

Substitute $40\text{V}$ for $v_1$ in equation (4).

$\begin{array}{l}-0.5\left(40\right)+0.625v_2=35\\ -20+0.625v_2=35\\ 0.625v_2=35+20\\ 0.625v_2=55\end{array}$

Simplify the equation as follows.

$\begin{array}{c}{v}_2=\frac{55}{0.625}\\ =88\text{V}\end{array}$

Write the expression for power dissipated through $1\Omega$ resistor.

$P_{1\Omega }=\frac{{\left(v_1-0\right)}^2}{1\Omega }$ (5)

Substitute $40\text{V}$ for $v_1$ in equation (5) to find $P_{1\Omega }$.

$\begin{array}{c}{P}_{1\Omega }=\frac{{\left(40\text{V}-0\right)}^2}{1\Omega }\\ =1600\text{W}\end{array}$

Write the expression for power dissipated through $2\Omega$ resistor.

$P_{2\Omega }=\frac{{\left(v_2-0\right)}^2}{2\Omega }$ (6)

Substitute $88\text{V}$ for $v_2$ in equation (6) to find $P_{2\Omega }$.

$\begin{array}{c}{P}_{2\Omega }=\frac{{\left(88\text{V}-0\right)}^2}{2\Omega }\\ =\frac{7744{\text{V}}^{\text{2}}}{2\Omega }\\ =3872\text{W}\end{array}$

Write the expression for power dissipated through $4\Omega$ resistor.

$P_{4\Omega }=\frac{{\left(v_1-280\right)}^2}{4\Omega }$ (7)

Substitute $40\text{V}$ for $v_1$ in equation (7) to find $P_{4\Omega }$.

$\begin{array}{c}{P}_{4\Omega }=\frac{{\left(40\text{V}-280\text{V}\right)}^2}{4\Omega }\\ =\frac{{\left(-240\text{V}\right)}^2}{4\Omega }\\ =\frac{57600{\text{V}}^{\text{2}}}{4\Omega }\\ =14400\text{W}\end{array}$

Write the expression for power dissipated through $8\Omega$ resistor.

$P_{8\Omega }=\frac{{\left(v_2-280\text{V}\right)}^2}{8\Omega }$ (8)

Substitute $88\text{V}$ for $v_2$ in equation (8) to find $P_{8\Omega }$.

$\begin{array}{c}{P}_{8\Omega }=\frac{{\left(88\text{V}-280\text{V}\right)}^2}{8\Omega }\\ =\frac{{\left(-192\text{V}\right)}^2}{8\Omega }\\ =\frac{36864{\text{V}}^{\text{2}}}{8\Omega }\\ =4608\text{W}\end{array}$

Conclusion:

Therefore, the power dissipated in $1\Omega$, $2\Omega$, $4\Omega$, and $8\Omega$ resistors are $1600\text{W}$, $3872\text{W}$, $14400\text{W}$, and $4608\text{W}$ respectively.