Chapter 2, Problem 1P

a.

To determine

Explain whether the interconnection of ideal sources in the circuit in Figure P2.1 in the textbook is valid or not.

Answer to Problem 1P

The interconnection of ideal sources in the given circuit is valid.

Explanation of Solution

Given data:

Refer to Figure P2.1 in the textbook for required data.

Description:

All the sources in the given circuit are independent sources. The independent voltage source can carry any current that required by the connection and the independent current source can support any voltage that required by the connection.

As the given circuit consists of all ideal independent sources and the Kirchhoff’s law is not violated, the interconnection of ideal sources in the given circuit is valid.

Conclusion:

Thus, the interconnection of ideal sources in the given circuit is valid.

b.

To determine

Identify the power developing sources and power absorbed sources in the given circuit.

Answer to Problem 1P

The power developing sources are 15 V voltage source and 5 A current source, and the power absorbing source is 20 V voltage source.

Explanation of Solution

Formula used:

Write the expression for power developed or absorbed by the source (voltage or current) as follows:

$p=v_s{i}_s$ (1)

Here,

$i_s$ is the current through the source and

$v_s$ is the voltage across the source.

Calculation:

If the current enters from the negative terminal and leaves from the positive terminal of a voltage source then the source delivers the power. The delivering power is obtained with negative sign.

If the current enters from the positive terminal and leaves from the negative terminal of a voltage source then the source absorbs the power. The absorbing power is obtained with positive sign.

From the given circuit, current of 5 A enters from the positive terminal of 20 V source. Therefore, 20 V voltage source absorbs the power.

Rewrite the expression in Equation (1) to find the power absorbed by the 20 V voltage source as follows:

$p_{20\text{V}}={\left(v_s\right)}_{20\text{V}}{\left(i_s\right)}_{20\text{V}}$

From given circuit, the values of ${\left(v_s\right)}_{20\text{V}}$ and ${\left(i_s\right)}_{20\text{V}}$ are 20 V and 5 A respectively. Substitute 20 V for ${\left(v_s\right)}_{20\text{V}}$ and 5 A for ${\left(i_s\right)}_{20\text{V}}$ to find the power absorbed by the 20 V voltage source.

$\begin{array}{c}{p}_{20\text{V}}=\left(20\text{V}\right)\left(5\text{A}\right)\\ =100\text{VA}\\ =100\text{W}\left\{\because \text{1}\text{VA=1}\text{W}\right\}\end{array}$

Therefore, the power absorbed by the 20 V voltage source is 100 W.

From the given circuit, current of 5 A enters from the negative terminal of 15 V source. Therefore, 15 V voltage source develops the power.

Rewrite the expression in Equation (1) to find the power developed by the 15 V voltage source as follows:

$p_{15\text{V}}={\left(v_s\right)}_{15\text{V}}{\left(i_s\right)}_{15\text{V}}$

From given circuit, the values of ${\left(v_s\right)}_{15\text{V}}$ and ${\left(i_s\right)}_{15\text{V}}$ are $-15\text{V}$ and 5 A respectively. Substitute $-15\text{V}$ for ${\left(v_s\right)}_{15\text{V}}$ and 5 A for ${\left(i_s\right)}_{15\text{V}}$ to find the power developed by the 15 V voltage source.

$\begin{array}{c}{p}_{15\text{V}}=\left(-15\text{V}\right)\left(5\text{A}\right)\\ =-75\text{W}\end{array}$

Therefore, the power developed by the 15 V voltage source is $-75\text{W}$. The negative sign indicates the developed power by the source.

From the given circuit, the voltage drop across 5 A current source is calculated by using KVL to the circuit as follows:

$\begin{array}{l}+20\text{V}-15\text{V}+v_{\text{5}\text{A}}=0\\ v_{\text{5}\text{A}}=-5\text{V}\end{array}$

Therefore, the voltage drop across 5 A current source is $-5\text{V}$. That is, current 5 A enters the negative terminal and leaves from the positive terminal of $-5\text{V}$ voltage drop. Therefore, 5 A current source develops the power.

Rewrite the expression in Equation (1) to find the power developed by the 5 A current source as follows:

$p_{\text{5}\text{A}}={\left(v_s\right)}_{\text{5}\text{A}}{\left(i_s\right)}_{\text{5}\text{A}}$

From given circuit, the values of ${\left(v_s\right)}_{\text{5}\text{A}}$ and ${\left(i_s\right)}_{\text{5}\text{A}}$ are $-5\text{V}$ and 5 A respectively. Substitute $-5\text{V}$ for ${\left(v_s\right)}_{\text{5}\text{A}}$ and 5 A for ${\left(i_s\right)}_{\text{5}\text{A}}$ to find the power developed by the 5 A current source.

$\begin{array}{c}{p}_{\text{5}\text{A}}=\left(-5\text{V}\right)\left(5\text{A}\right)\\ =-25\text{W}\end{array}$

The negative sign indicates the developed power by the source. Therefore, the power developed by the 5 A current source is 25 W.

Conclusion:

Thus, the power developing sources are 15 V voltage source and 5 A current source, and the power absorbing source is 20 V voltage source.

c.

To determine

Verify the power absorbed in the circuit is equal to the power developed in the circuit.

Answer to Problem 1P

The power absorbed in the circuit is equal to the power developed in the circuit.

Explanation of Solution

Calculation:

From Part (b), write the expression for power absorbed in the circuit as follows:

${\displaystyle \sum p_{abs}}=p_{20\text{V}}$

Substitute 100 W for $p_{20\text{V}}$ to obtain the power absorbed in the circuit.

${\displaystyle \sum p_{abs}}=100\text{W}$

From Part (b), the power developed in the circuit as follows:

${\displaystyle \sum p_{del}}=p_{15\text{V}}+p_{\text{5}\text{A}}$

Substitute 75 W for $p_{15\text{V}}$ and 25 W for $p_{\text{5}\text{A}}$ to obtain the power developed in the circuit.

$\begin{array}{c}{\displaystyle \sum p_{del}}=75\text{W}+25\text{W}\\ =100\text{W}\end{array}$

From the calculation, it is clear that the power absorbed in the circuit is equal to the power developed in the circuit.

Conclusion:

Thus, the power absorbed in the circuit is equal to the power developed in the circuit.

d.

To determine

Explain whether the interconnection of ideal sources in the modified circuit is valid or not and identify the power developing and absorbing sources in the modified circuit. Verify the power absorbed in the modified circuit is equal to the power delivered in the circuit.

Answer to Problem 1P

The interconnection of ideal sources in the modified circuit is valid, voltage source of 20 V and voltage source of 15 V are power developing sources and 5 A current source is a power absorbing source. The power absorbed in the modified circuit is equal to the power developed in the circuit.

Explanation of Solution

Description:

Redraw the circuit by reversing the polarity of 20 V voltage source as shown in Figure 1.

As the modified circuit in Figure 1 consists of all ideal independent sources, the interconnection of ideal sources in the modified circuit is valid.

From Figure 1, current of 5 A enters from the negative terminal of 20 V source. Therefore, 20 V voltage source develops the power.

Rewrite the expression in Equation (1) to find the power developed by the 20 V voltage source as follows:

$p_{20\text{V}}={\left(v_s\right)}_{20\text{V}}{\left(i_s\right)}_{20\text{V}}$

From given circuit, the values of ${\left(v_s\right)}_{20\text{V}}$ and ${\left(i_s\right)}_{20\text{V}}$ are $-20\text{V}$ and 5 A respectively. Substitute $-20\text{V}$ for ${\left(v_s\right)}_{20\text{V}}$ and 5 A for ${\left(i_s\right)}_{20\text{V}}$ to find the power delivered by the 20 V voltage source.

$\begin{array}{c}{p}_{20\text{V}}=\left(-20\text{V}\right)\left(5\text{A}\right)\\ =-100\text{VA}\\ =-100\text{W}\end{array}$

Therefore, the power delivered by the 20 V voltage source is 100 W.

From the given circuit, current of 5 A enters from the negative terminal of 15 V source. Therefore, 15 V voltage source develops the power.

Rewrite the expression in Equation (1) to find the power developed by the 15 V voltage source as follows:

$p_{15\text{V}}={\left(v_s\right)}_{15\text{V}}{\left(i_s\right)}_{15\text{V}}$

From given circuit, the values of ${\left(v_s\right)}_{15\text{V}}$ and ${\left(i_s\right)}_{15\text{V}}$ are $-15\text{V}$ and 5 A respectively. Substitute $-15\text{V}$ for ${\left(v_s\right)}_{15\text{V}}$ and 5 A for ${\left(i_s\right)}_{15\text{V}}$ to find the power developed by the 15 V voltage source.

$\begin{array}{c}{p}_{15\text{V}}=\left(-15\text{V}\right)\left(5\text{A}\right)\\ =-75\text{W}\end{array}$

Therefore, the power developed by the 15 V voltage source is 75 W.

From the given circuit, the voltage drop across 5 A current source is calculated by using KVL to the circuit as follows:

$\begin{array}{l}-20\text{V}-15\text{V}+v_{\text{5}\text{A}}=0\\ v_{\text{5}\text{A}}=35\text{V}\end{array}$

Therefore, the voltage drop across 5 A current source is 35 V. That is, current 5 A enters the positive terminal and leaves from the negative terminal of 35 V voltage drop. Therefore, 5 A current source absorbs the power.

Rewrite the expression in Equation (1) to find the power absorbed by the 5 A current source as follows:

$p_{\text{5}\text{A}}={\left(v_s\right)}_{\text{5}\text{A}}{\left(i_s\right)}_{\text{5}\text{A}}$

From given circuit, the values of ${\left(v_s\right)}_{\text{5}\text{A}}$ and ${\left(i_s\right)}_{\text{5}\text{A}}$ are 35 V and 5 A respectively. Substitute 35 V for ${\left(v_s\right)}_{\text{5}\text{A}}$ and 5 A for ${\left(i_s\right)}_{\text{5}\text{A}}$ to find the power absorbed by the 5 A current source.

$\begin{array}{c}{p}_{\text{5}\text{A}}=\left(35\text{V}\right)\left(5\text{A}\right)\\ =175\text{W}\end{array}$

Therefore, the power absorbed by the 5 A current source is 175 W.

From the calculation, write the expression for power absorbed in the circuit as follows:

${\displaystyle \sum p_{abs}}=p_{\text{5}\text{A}}$

Substitute 175 W for $p_{\text{5}\text{A}}$ as follows:

${\displaystyle \sum p_{abs}}=175\text{W}$

From the calculation, the power developed in the circuit as follows.

${\displaystyle \sum p_{del}}=p_{20\text{V}}+p_{15\text{V}}$

Substitute 100 W for $p_{20\text{V}}$ and 75 W for $p_{15\text{V}}$ to obtain the power developed in the circuit.

$\begin{array}{c}{\displaystyle \sum p_{del}}=100\text{W}+75\text{W}\\ =175\text{W}\end{array}$

From the calculation, it is clear that the power absorbed in the circuit is equal to the power developed in the circuit.

Conclusion:

Thus, the interconnection of ideal sources in the modified circuit is valid, voltage source of 20 V and voltage source of 15 V are power developing sources and 5 A current source is a power absorbing source. The power absorbed in the modified circuit is equal to the power developed in the circuit.