Chapter 3, Problem 34E

Although drawn so that it may not appear obvious at first glance, the circuit of Fig. 3.74 is in fact a single-node-pair circuit, (a) Determine the power absorbed by each resistor, (b) Determine the power supplied by each current source, (c) Show that the sum of the absorbed power calculated in (a) is equal to the sum of the supplied power calculated in (b).

FIGURE 3.74

To determine

Find power absorbed by each resistor.

Answer to Problem 34E

Power absorbed by the resistor $R_1$ is $345.3\mu \text{W}$, power absorbed by resistor $R_2$ is $579.7\mu \text{W}$ and power absorbed by resistor $R_3$ is $1.623\text{mW}$.

Explanation of Solution

Calculation:

The circuit diagram is redrawn as shown in Figure 1.

Refer to the redrawn Figure 1.

The expression for KCL at node $A$ is as follows.

$i_1+\frac{v_A}{R_3}+\frac{v_A}{R_1}+\frac{v_A}{R_2}-i_2=0$ (1)

Here,

$i_1$ and $i_2$ are the currents in the circuit,

$v_A$ is the voltage at node $A$ and

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

The expression for power absorbed by resistor is as follows.

$p=\frac{v^2}{R}$ (2)

Here,

$p$ is power absorbed by resistor,

$v$ is the voltage and

$R$ is value of resistance.

Refer to the redrawn Figure 1.

Substitute $5\text{mA}$ for $i_1$, $3\text{mA}$ for $i_2$, $4.7\text{kΩ}$ for $R_1$, $2.8\text{kΩ}$ for $R_2$ and $1\text{kΩ}$ for $R_3$ in equation (1).

$\begin{array}{l}5\text{mA}+\frac{v_A}{1\text{kΩ}}+\frac{v_A}{4.7\text{kΩ}}+\frac{v_A}{2.8\text{kΩ}}-3\text{mA}=0\\ 2\text{mA}+10\times {10}^{-4}{v}_A+2.127\times {10}^{-4}{v}_A+3.571\times {10}^{-4}{v}_A=0\\ 15.7\times {10}^{-4}{v}_A+2\times {10}^{-3}=0\left\{\because \text{ 1 mA}={\text{10}}^{-3}\text{ A}\right\}\end{array}$$15.7\times {10}^{-4}{v}_A+2\times {10}^{-3}=0$ (3)

Rearrange equation (3) for $v_A$.

$v_A=-1.274\text{V}$

Substitute $-1.274\text{V}$ for $v$ and $4.7\text{kΩ}$ for $R$ in equation (2).

$\begin{array}{c}p=\frac{{\left(-1.274\text{V}\right)}^2}{4.7\text{kΩ}}\\ =\frac{1.6231{\text{V}}^2}{4.7\times {10}^3\text{Ω}}\left\{\because 1\text{kΩ}={10}^3\text{Ω}\right\}\\ =0.3453\times {10}^{-3}\text{W}\\ =\text{345}\text{.3}\times {10}^{-6}\text{W}\end{array}$

Simplify for $p$.

$p=345.3\mu \text{W}\left\{\because {10}^{-6}\text{W}=\text{1}\mu \text{W}\right\}$

So, power absorbed by resistor $R_1$ is $345.3\mu \text{W}$.

Substitute $-1.274\text{V}$ for $v$ and $2.8\text{kΩ}$ for $R$ in equation (2).

$\begin{array}{c}p=\frac{{\left(-1.274\text{V}\right)}^2}{2.8\text{kΩ}}\\ =\frac{1.6231{\text{V}}^2}{2.8\times {10}^3\text{Ω}}\left\{\because 1\text{kΩ}={10}^3\text{Ω}\right\}\\ =0.5797\times {10}^{-3}\text{W}\\ =579.\text{7}\times {10}^{-6}\text{W}\end{array}$

Simplify for $p$.

$p=579.7\mu \text{W}\left\{\because {10}^{-6}\text{W}=\text{1}\mu \text{W}\right\}$

So, the power absorbed by resistor $R_2$ is $579.7\mu \text{W}$.

Substitute $-1.274\text{V}$ for $v$ and $1\text{kΩ}$ for $R$ in equation (2).

$\begin{array}{c}p=\frac{{\left(-1.274\text{V}\right)}^2}{1\text{kΩ}}\\ =\frac{1.623{\text{V}}^2}{1\times {10}^3\text{Ω}}\left\{\because 1\text{kΩ}={10}^3\text{Ω}\right\}\\ =1.623\times {10}^{-3}\text{W}\\ =1.623\text{mW}\left\{\because {10}^{-3}\text{W}=\text{1}\text{mW}\right\}\end{array}$

So, the power absorbed by resistor $R_3$ is $1.623\text{mW}$.

Conclusion:

Thus, the power absorbed by resistor $R_1$ is $345.3\mu \text{W}$, power absorbed by resistor $R_2$ is $579.7\mu \text{W}$ and power absorbed by resistor $R_3$ is $1.623\text{mW}$.

To determine

Find the power supplied by each current source.

Answer to Problem 34E

Power supplied by dependent current source $i_1$ is $\text{6}\text{.37}\text{mW}$ and power supplied by dependent current source $i_2$ is $\text{3}\text{.833}\text{mW}$.

Explanation of Solution

Formula used:

The expression for power supplied by current source is as follows.

$p=vi$ (4)

Here,

$p$ is the power supplied by current source,

$i$ is the current and

$v$ is the voltage.

Calculation:

Refer to the redrawn Figure 1.

As current direction for independent current source $i_1$ is downward and we are calculating power supply by this so we will use value of magnitude with negative sign means value of $i_1$ is $-5\text{mA}$.

Substitute $-1.274\text{V}$ for $v$ and $-5\text{mA}$ for $i$ in equation (4).

$\begin{array}{c}p=\left(-1.274\text{V}\right)\left(-5\text{mA}\right)\\ =\left(-1.274\text{V}\right)\left(-5\times {10}^{-3}\text{A}\right)\left\{\because 1\text{mA}={10}^{-3}\text{A}\right\}\\ =6.37\times {10}^{-3}\text{W}\\ =\text{6}\text{.37}\text{mW}\left\{\because {10}^{-3}\text{W}=1\text{mW}\right\}\end{array}$

So power supplied by dependent current source $i_1$ is $\text{6}\text{.37}\text{mW}$.

Substitute $-1.274\text{V}$ for $v$ and $3\text{mA}$ for $i$ in equation (4).

$\begin{array}{c}p=\left(-1.274\text{V}\right)\left(3\text{mA}\right)\\ =\left(-1.274\text{V}\right)\left(3\times {10}^{-3}\text{A}\right)\left\{\because 1\text{mA}={10}^{-3}\text{A}\right\}\\ =-3.833\times {10}^{-3}\text{W}\\ =-\text{3}\text{.833}\text{mW}\left\{\because {10}^{-3}\text{W}=1\text{mW}\right\}\end{array}$

So power supplied by dependent current source $i_2$ is $-\text{3}\text{.833}\text{mW}$.

Conclusion:

Thus, the power supplied by dependent current source $i_1$ is $\text{6}\text{.37}\text{mW}$ and power supplied by dependent current source $i_2$ is $-\text{3}\text{.833}\text{mW}$.

To determine

Verify that sum of power absorbed and sum of power supplied in the circuit is same.

Answer to Problem 34E

Sum of power absorbed and sum of power supplied in the circuit is same.

Explanation of Solution

Formula used:

The expression for power is as follows.

$p=p_1+p_2+p_3$ (5)

Here,

$p$ is the total power and

$p_1$, $p_2$ and $p_3$ are power.

Calculation:

Substitute $345.3\mu \text{W}$ for $p_1$, $579.7\mu \text{W}$ for $p_2$ and $1.623\text{mW}$ for $p_3$ in equation (5).

$\begin{array}{c}p=345.3\mu \text{W}+579.7\mu \text{W}+1.623\text{mW}\\ =345.3\text{+W}+579.7\mu \text{W}+1623\mu \text{W}\left\{\because {\text{10}}^{-3}\text{mW}=\text{1}\mu \text{W}\right\}\\ =2548\mu \text{W}\end{array}$

So, the total power absorbed is $2548\mu \text{W}$.

Substitute $\text{6}\text{.37}\text{mW}$ for $p_1$ and $-\text{3}\text{.833}\text{mW}$ for $p_2$ in equation (5).

$\begin{array}{c}p=p_1+p_2\\ =\text{6}\text{.37}\text{mW}+\left(-\text{3}\text{.833}\text{mW}\right)\\ =2.538\text{mW}\\ =\text{2548}\mu \text{W}\left\{\because {\text{10}}^{-3}\text{mW}=\text{1}\mu \text{W}\right\}\end{array}$

So total power supplied is $2548\mu \text{W}$.

Conclusion:

Thus, sum of power absorbed and sum of power supplied in the circuit is same.