Solved Problems in Electrical Circuits

Total Energy Stored - Circuit with Capacitors and Inductors

admin — Fri, 19 Nov 2010 04:42:19 +0000

Find the total energy stored in the circuit.

Fig. (1-28-1) - The circuit

Solution
The circuit contains only dc sources. Recall that an inductor is a short circuit to dc and a capacitor is an open circuit to dc. These can be easily verified from their current-voltage characteristics. For an inductor, we have

$v(t)=L \frac{d i(t)}{dt}$ . Since a dc current does not vary with time,

$\frac{d i(t)}{dt}=0$ . Hence, the voltage across the inductor is zero for any dc current. This is to say that dc current passes through the inductor without any voltage drop, exactly similar to a short circuit. For a capacitor, the current-voltage terminal characteristics is

$i(t)=L \frac{d v(t)}{dt}$ . Voltage drop across passive elements due to dc currents does not vary with time. Therefore,

$\frac{d v(t)}{dt}=0$ and consequently the current of the capacitor is zero. This is to say that dc current does not pass through the capacitor regardless of the voltage amount. This is similar to the behavior of an open circuit. Please note that unlike dc current, ac current passes through capacitors in general.

To find dc current of the inductors and dc voltage drop across the capacitors, we replace them with their equivalent elements, short circuits and open circuits, respectively, as shown in Fig. (1-28-2).

Fig. (1-28-2) - Replacing inductors and capacitors with their dc equivalents

It is easy to find that

$I_{1\Omega}=\frac{9V}{2\Omega +1 \Omega}=3A$ and

$I_{3mH}=1A$ . Therefore,

$I_{2mH}=I_{3mH}+I_{1\Omega}=4A$ ,

$V_{20\mu F}=-1\Omega \times I_{1\Omega}=-3V$ and

$V_{10\mu F}=-2\Omega \times I_{1\Omega}=-6V$ . The total energy stored in the circuit is the sum of the energy stored in elements capable of storing energy, i.e. two capacitors and two inductors. Recall that the energy stored in an inductor is

$w_{L}=\frac{1}{2}Li^2_L(t)$ and is equal to

$w_{C}=\frac{1}{2}CV^2_C(t)$ for a capacitor. Thus,

$w_{3mH}=\frac{1}{2}(3mH)(1A)^2=1.5\,mJ$

$w_{2mH}=\frac{1}{2}(2mH)(4A)^2=16\,mJ$

$w_{20 \mu F}=\frac{1}{2}(20 \mu F)(-3V)^2=90\,\mu J$

$w_{10 \mu F}=\frac{1}{2}(10 \mu F)(-6V)^2=180\,\mu J$
The total stored energy is

$17.77 mJ$ .

Now, switch the sources as shown in Fig. (1-28-3) and calculate the total stored energy. The answer is $60.655 mJ$ . Some of the key quantities are $I_{3mH}=4.5A$ , $I_{2mH}=5.5A$ , $V_{10\mu F}=2V$ and $V_{20\mu F}=1V$ .

Fig. (1-28-3) - Homework

Thévenin's Theorem - Circuit with Two Independent Sources

admin — Tue, 16 Nov 2010 06:25:05 +0000

Use Thévenin's theorem to determine $I_O$ .

Fig. (1-27-1) - Circuit with two independent sources

Solution
Lets break the circuit at the

$3\Omega$ load as shown in Fig. (1-27-2).

Fig. (1-27-2) - Breaking circuit at the load

Now, we should find an equivalent circuit that contains only an independent voltage source in series with a resistor, as shown in Fig. (1-27-3).

Fig. (1-27-3) - The Thevenin equivalent circuit

Unknowns are

$V_{Th}$ and

$R_{Th}$ .

$V_{Th}$ is the open circuit voltage

$V_{OC}$ shown in Fig. (1-27-2).
It is trivial that the current of

$2\Omega$ resistor is equal to the current of the current source, i.e.

$I_{2\Omega}=-1A$ . Therefore,

$V_{OC}=V_{2\Omega}=2\Omega \times I_{2\Omega}=-2V$ . The Thévenin theorem says that

$V_{Th}=V_{OC}=-2V$ . Please note that it is not saying that

$V_{OC}$ is the voltage across the load in the original circuit (Fig. (1-27-1)).

To find the other unknown,

$R_{Th}$ , we turn off independent sources and find the equivalent resistance seen from the port, as this is an easy way to find

$R_{Th}$ for circuits without dependent sources. Recall that in turning independent sources off, voltage sources should be replace with short circuits and current sources with open circuits. By turning sources off, we reach at the circuit shown in Fig. (1-27-4).

Fig. (1-27-4) - Turning off the sources to find Rth

The

$6\Omega$ resistor is short circuited and the

$5\Omega$ one is open. Therefore, their currents are zero and

$R_{Th}=2\Omega$ .
Now that we have found

$V_{Th}$ and

$R_{Th}$ , we can calculate

$I_O$ in the original circuit shown in Fig. (1-27-1) using the Thévenin equivalent circuit depicted in Fig. (1-27-3). It is trivial that

$I_{O}=\frac{V_{th}}{R_{Th}+3 \Omega}=\frac{-2V}{2\Omega+3 \Omega}=-\frac{2}{5}A$ .

We used the Thévenin Theorem to solve this circuit. A much more easier way to find $I_O$ here is to use the current devision rule. The current of the current source is divided between $2\Omega$ and $3\Omega$ resistors. Therefore,
$I_{O}=\frac{2\Omega}{2\Omega+3 \Omega} \times (-1A)=-\frac{2}{5}A$

Now, replace the current source with a $-1V$ voltage source as shown below and solve the problem. The answers are $V_{th}=\frac{4}{7} V$ , $R_{th}=\frac{10}{7}\Omega$ and $I_O=\frac{4}{31}A$ . Please let me know how it goes and leave me a comment if you need help

Fig (1-27-5) - Homework

Thévenin's Theorem - Circuit with An Independent Source

admin — Thu, 28 Oct 2010 06:07:56 +0000

Use Thévenin's theorem to determine $V_O$ .

Fig. (1-26-1) - The Circuit

Solution
To find the Thévenin equivalent, we break the circuit at the

$4\Omega$ load as shown below.

Fig. (1-26-2) - Breaking the circuit at the load

So, our goal is to find an equivalent circuit that contains only an independent voltage source in series with a resistor, as shown in Fig. (1-26-3), in such a way that the current-voltage relationship at the load is not changed.

Fig. (1-26-3) - Replacing the Thevenin equivalent circuit

Now, we need to find

$V_{Th}$ and

$R_{Th}$ .

$V_{Th}$ is equal to the open circuit voltage

$V_{OC}$ shown in Fig. (1-26-2). The current of

$2\Omega$ resistor is zero because one of its terminals is not connected to any element; therefore, current cannot pass through it. Since the current of

$2\Omega$ resistor is zero, the

$9V$ voltage source,

$3\Omega$ and

$6\Omega$ resistors form a voltage divider circuit and the voltage across the

$6\Omega$ resistor can be determined by the voltage devision rule. Please not that we are able to use the voltage devision rule here just because the current of the

$2\Omega$ resistor is zero. You may ask that there is no reason to prove that the current of the

$2\Omega$ resistor is zero in the original circuit shown in Fig. (1-26-1). That is correct. However, we are calculating

$V_{OC}$ for the circuit shown in Fig. (1-26-1) and this is a different circuit. The Thévenin theorem guarantees that

$V_{Th}=V_{OC}$ , it is not saying that

$V_{OC}$ is the voltage across the load in the original circuit.

$V_{6\Omega}=\frac{6\Omega}{3\Omega+6\Omega}\times 9 V= 6V$
Since the current of the

$2\Omega$ resistor is zero:

$V_{OC}=V_{6\Omega}=6V$

$V_{Th}=V_{OC}=6V$
Now, we need to find

$R_{Th}$ . An easy way to find

$R_{Th}$ for circuits without dependent sources is to turn off independent sources and find the equivalent resistance seen from the port. Recall that voltage sources should be replace with short circuits and current sources with open circuits. Here, there is only a voltage source that should be replaced by short circuit as shown in Fig. (1-26-4).

Fig. (1-26-4) - Turning off the voltage source to find Rth

It is trivial to see that the

$3 \Omega$ and

$6 \Omega$ resistors are connected in parallel and then wired in series to the

$2\Omega$ resistor. Therefore,

$R_{Th}=(3\Omega || 6\Omega)+2\Omega=\frac{3\Omega \times 6\Omega}{3\Omega + 6\Omega}+2\Omega=4\Omega$ .
Now that

$V_{Th}$ and

$R_{Th}$ are found, we can use the Thévenin equivalent circuit depicted in Fig. (1-26-3) to calculate

$V_O$ in the original circuit shown in Fig. (1-26-1). The voltage devision rule can be used here to find

$V_O$ . We have,

$V_{O}=\frac{4\Omega}{R_{Th}+4 \Omega}\times V_{Th}=\frac{4\Omega}{4\Omega+4 \Omega}\times 6V=3V$ .

Superposition Method - Circuit With Dependent Sources

admin — Wed, 08 Sep 2010 05:49:51 +0000

Determine $I_x$ , $I_y$ and $V_z$ using the superposition method.

Solution
I. Contribution of the $-2V$ voltage source:
We need to turn off the current source by replacing it with an open circuit. Recall that we do not turn off dependent sources. The resulting circuit is shown below.

In the left hand side loop, we have
$6I_{x1}=1 \Omega \times I_{y1}$

and in the right hand loop, it is trivial that
$2I_{y1}=I_{x1}$ .
Therefore, $I_{x1}=I_{y1}=0$ . Applying KVL around the inner loop,
$-V_{z1}-1\Omega\times I_{y1}+(-2V)-2\Omega \times I_{x1}=0$
Substituting $I_{x1}=I_{y1}=0$ , we have
$V_{z1}=-2V$ .

II. Contribution of the $11A$ current source:
The independent voltage source must be replaced with a short circuit as shown below.

The $1\Omega$ resistor is parallel with the dependent voltage source, Therefore, $V_{1\Omega}=6I_{x2}$ and since $V_{1\Omega}=1\Omega \times I_{y2}$ , we have $6I_{x2}=I_{y2}$ . Applying KCL at the right bottom node, $11A-I_{x2}+2I_{y2}$ . So,

$\left\{ \begin{array}{l} 6I_{x2}=I_{y2} \\ I_{x2}-2I_{y2}=11 \end{array} \right. \to \left\{ \begin{array}{l} I_{x2}=-1 A \\ I_{y2}=-6A \end{array} \right.$
Applying KVL around the inner loop,
$+2\Omega \times I_{x2}+1 \Omega \times I_{y2}+V_{z2}=0 \to V_{z2}=8V.$

III. The final result

$I_x=I_{x1}+I_{x2}=-1 A$
$I_y=I_{y1}+I_{y2}=-6 A$
$V_z=V_{z1}+V_{z2}=6V$

Superposition Problem with Four Voltage and Current Sources

admin — Tue, 07 Sep 2010 06:33:25 +0000

Determine $V_x$ and $I_x$ using the superposition method.

Solution
I. Contribution of the $-5V$ voltage source:

To find the contribution of the $-5V$ voltage source, other three sources should be turned off. The $3V$ voltage source should be replaced by short circuit. The current source should be replaced with open circuits, as shown below.

It is trivial that $I_{x1}= \frac{-5 V}{2 \Omega}=-2.5 A$ . The current of the $3\Omega$ resistor is zero. Using KVL, $-(-5V)+V_{3\Omega}-V_{x1}=0 \to V_{x1}=-(-5V)=5V$ .

II. Contribution of the $3V$ voltage source:
Similarly, the $-5V$ voltage source becomes a short circuit and the current source should be replaced with open circuits:

The current of the $2\Omega$ resistor is zero because of being short circuited. It is trivial that $I_{x2}= 0 A$ (current of an open circuit). The current of the $3\Omega$ resistor is also zero. Using KVL, $-(3V)+V_{2\Omega}+V_{x2}+V_{3\Omega}=0 \to V_{x2}=3V$ .

III. Contribution of the $-1A$ current source:
The voltage sources should be replaced by short circuits and the $2A$ current source becomes with open circuit:

Again, the $2\Omega$ resistor is short circuited and its current is zero. it is clear that $I_{x3}= 1 A$ . The current of the $3\Omega$ resistor is equal to $-1A$ . Using KVL, $V_{x3}+V_{3\Omega}=0 \to V_{x3}+(-1A)\times (3\Omega)=0 \to V_{x3}=3V$ .

IV. Contribution of the $2A$ current source:
Likewise, the voltage sources should be replaced by short circuits and the $-1A$ current source becomes with open circuit:

Again, the $2\Omega$ resistor is short circuited and its current is zero. it is also trivial that $I_{x4}= 0A$ . The current of the $3\Omega$ resistor is $2A$ . Using KVL, $V_{x4}+V_{3\Omega}=0 \to V_{x4}+(2A)\times (3\Omega)=0 \to V_{x4}=-6V$ .

V. Adding up the individual contributions algebraically:

$V_x=V_{x1}+V_{x2}+V_{x3}+V_{x4}=5V+3V+3V-6V\to V_x=5V$
$I_x=I_{x1}+I_{x2}+I_{x3}+I_{x4}=-2.5A+1A+0A-0A\to I_x=-1.5A$

Turning Sources Off

admin — Thu, 02 Sep 2010 05:40:36 +0000

Turning off a source, which is usually used in solving circuits with superposition method, means setting its value equal to zero. For a voltage source, setting the voltage equal to zero means that it produces zero voltage between its terminals. Therefore, the voltage source must insure that the voltage across two terminals is zero. Replacing the source with a short circuit can do that. Thus, voltage sources become a short circuit when turned off.

For a current source, setting the current equal to zero means that it produces zero current. Therefore, the current source must insure that no current flows through its branch. An open circuit can do that. Hence, to turn off a current source it should be replaced by an open circuit.

How about dependent sources? The voltage/current of a dependent source is dependent on other variables of the circuit. Therefore, dependent sources cannot be turned off.

Example I: Turn off sources one by one.

Example 1

Solution:
I) The voltage source:

Turning off the voltage source

II) The current source:

Turning off the current source

Example 2: For each source, leave the source on and turn off all other sources.

Example 2

Solution
I. $V_1$ :

Contribution of V_1

II. $V_2$ :

Contribution of V2

III.

$I_1$ :

Contribution of I1

IV. $I_2$ :

Contribution of I2

Example 3: For each source, leave the source on and turn off all other sources.

Example 3

Solution
I.

$V_1$ :

Contribution of V1

II. $I_1$

Contribution of I1

Recall that dependent sources cannot be turned off.

Electrical Circuits eBooks

admin — Mon, 05 Jul 2010 02:05:03 +0000

Nodal Analysis eBook

You are welcome to download our free ebook about Nodal Analysis from here. ~~You need to sign up for our site before downloading.~~ Please note that we will never sell, trade or share your email with a third-party. Occasionally, we will send you an email about new posts, ebooks and exciting things happening in solved-problems.com.

content

Introduction
Reference Node
Node Voltages
Nodal Analysis Steps
Complicated Cases
- Circuits with Non-grounded Voltage Sources
- Circuits with Dependent Current Sources
- Circuits with Dependent Voltage Sources
More Problems

Nodal Analysis

admin — Fri, 02 Jul 2010 04:12:35 +0000

Nodal Analysis eBook

You are welcome to download our free ebook about Nodal Analysis from here. You need to sign up for our site before downloading. Please note that we will never sell, trade or share your email with a third-party. Occasionally, we will send you an email about new posts, ebooks and exciting things happening in solved-problems.com.

Nodal Analysis Steps

1) Identify all nodes in the circuit. Call the number of nodes N.

2) Select a reference node. Label it with reference (ground) symbol. As a general rule, the reference node is usually chosen to be

a node with largest number of elements connected to it, or
a node which is connected to the maximum number of voltage sources, or
a node of symmetry.
3) Assign a variable for each node whose voltage is unknown. If a voltage source is connected between a node and the reference node, the voltage is already known and it is not necessary to assign a variable. If there is a voltage source between two nodes, the difference between the node voltages equals to the voltage of the source. In this case, to reduce the number of unknowns assign a variable for one of the nodes and express the voltage of the other one with respect to the assigned variable.

4) If there are dependent sources in the circuit, write down equations that express their values in terms of node voltages.

5) Write down a KCL equation for each node by setting the total current flowing out of the node to zero. Recall that the KCL states that the algebraic sum of all currents entering and exiting a node is equal to zero. It is always a good idea to rearrange these equations into the form $A_1 \times V_1 + A_2 \times V_2 + \cdots + A_{N-1} \times V_{N-1} = C$ where $A_1, A_2, A_{N-1}$ and $C$ are some constants. If there are voltage sources between two unknown voltages, join the two nodes as a supernode. Note that you should have only one unknown variable for a supernode because the voltage of one the nodes can be expressed with respect to the voltage of the other one. For a supernode, the currents of the two nodes are combined in a single equation, and a new equation for the voltages is formed. For a circuit with N nodes and M voltage sources N - M - 1 simultaneous linearly independent equations can be written.

Here are some solved problems posted in solved-problems.com:

Solving a Circuit with Three Nodes by the Nodal Analysis

Solving by Nodal Analysis – Circuit with Four Nodes

Complicated Cases

The nodal analysis method is generally straightforward to apply, but becomes rather difficult in the following cases.

Non-grounded Voltage Sources

Since the current of a voltage source is independent of the voltage, it cannot be used in writing KCL equations. If one node of a voltage source is connected to the reference node, we do not need to know the current passing through the voltage source. The reason is that the voltage of the node can be easily determined by the voltage of the voltage source and there is no need to write KCL equation for the node.

Complicated cases are the ones where a voltage source is located between two non-reference nodes. In these cases, a supernode method should be used. A simple supernode is consist of a source and its nodes. In general, supernodes can have more than one voltage sources. After identifying a supernode, we need to define only one voltage variable for one of the nodes of the supernode and find the voltage of other node(s) with respect to that voltage variable. This equation relates node voltages of the supernode to each other. Then, we should treat a supernode as a node and write a KCL equation for all currents entering and leaving the super node. Now we have one equation and two unknowns (the node voltages). This equation should be added to the set of equations derived for other nodes and the new set of equations should be solved to determine all node voltages.

Check out this solved problem:

Nodal Analysis – Supernode

Dependent Current Source

When there is a dependent current source in the circuit, it should be treated as an independent current source but the variable which the current source depends on should be expressed in terms of node voltages. For example, if it is current of a resistor, Ohm's law should be used to state the variable in term of the node voltages of the resistor.

Here is a solved problem with a dependent current source:

Nodal Analysis – Dependent Current Source

Dependent Voltage Sources

A dependent voltage source can make the solution a bit challenging. The solution follows the same steps mentioned for dependent source with an extra step. After writing super-node KCL equation, the variable that the dependent source depends on should be written in terms of the node voltages.

The circuit of the following solved problem has a dependent voltage source:

Nodal Analysis – Dependent Voltage Source

Nodal Analysis - Circuit with Dependent Voltage Source

admin — Thu, 01 Jul 2010 22:53:22 +0000

Determine the power of each source after solving the circuit by the nodal analysis.

Answers: $P_{I_x}=0.497W, P_{1A}=-1.806W, P_{2A}=4.254W, P_{3V}=-3.87W,$ and $P_{5V}=-3.552W$

Solution

I. Identify all nodes in the circuit.
The circuit has 6 nodes as highlighted below.

II. Select a reference node. Label it with the reference (ground) symbol.

The right top node is connected to two voltage sources and has three elements. All other nodes also have three elements. Hence, we select the right top node because by this selection, we already know the node voltages of two other nodes, i.e. the ones that the reference node is connected to them by voltage sources.

III. Assign a variable for each node whose voltage is unknown.
We label the remaining nodes as shown above. Nodes of $V_3$ and $V_4$ are connected to the reference node through voltage sources. Therefore, $V_3$ and $V_4$ can be found easily by the voltages of the voltage sources. For $V_3$ , the negative terminal of the voltage source is connected to the node. Thus, $V_3$ is equal to minus the source voltage, $V_3=-5 V$ . The same argument applies to $V_4$ and $V_4=-3V$ .

IV. If there are dependent sources in the circuit, write down equations that express their values in terms of node voltages.

The voltage of the dependent voltage source is $I_x$ . We should find this value in terms of the node voltages. $I_x$ is the current of the $3\Omega$ - resistor. The voltage across the resistor is $V_2-V_4$ . We prefer to define $V_{3\Omega}$ as $V_2-V_4$ instead of $V_4-V_2$ to comply with passive sign convention. By defining $V_{3\Omega}$ as mentioned, $I_x$ is entering from the positive terminal of $V_{3\Omega}$ and we have $V_{3\Omega}= 3 \Omega \times I_x$ . Therefore, $I_x=\frac{V_2-V_4}{3\Omega}$ . $\to I_x=\frac{V_2}{3} +1$

V. Write down a KCL equation for each node.

Nodes of $V_1$ and $V_2$ :
These two nodes are connected through a voltage source. Therefore, they form a supernode and we can write the voltage of one in terms of the voltage of the other one. Please note that the voltage of the dependent voltage source is $I_x$ and we have $V_2=V_1+I_x$
$\to V_2=V_1+\frac{V_2}{3} +1 \to \frac{2}{3} V_2=V_1+1$
$\to V_1=\frac{2}{3} V_2-1$

KCL for the supernode:
$\frac{V_1-V_3}{5\Omega}+\frac{V_1-V_5}{2\Omega}+\frac{V_2-V_4}{3\Omega}-2A=0$
$\to \frac{V_1+5}{5}+\frac{V_1-V_5}{2}+\frac{V_2+3}{3}-2=0$
$\to \frac{V_1}{5}+\frac{V_1-V_5}{2}+\frac{V_2}{3}=0$
$\to 21V_1-15V_5+10V_2=0$
Substituting $V_1=\frac{2}{3} V_2-1$ ,
$\to 8V_2-5V_5=7$

Node of $V_5$ :
$\frac{V_5-V_1}{2 \Omega} +\frac{V_5-V_3}{1 \Omega}+1=0$
$\to -V_1 +3V_5-2V_3+2=0$
Substituting $V_1=\frac{2}{3} V_2-1$ and $V_3=-5 V$ ,
$\to -2 V_2 +9V_5=-39$

Here is the system of equations that we need to solve and obtain $V_2$ nd $V_5$ :

$\left\{ \begin{array}{l} I: 8V_2-5V_5=7 \\ II: -2 V_2 +9V_5=-39 \end{array} \right.$

We use elimination method to solve this system of equation:
$(II)\times 4 +(I): 31V_5=-149 \to$
$V_5=-4.806 V$
$V_2=\frac{9V_5+39}{2} \to$
$V_2=-2.127 V$
Using $V_1=\frac{2}{3} V_2-1$ ,
$V_1=-2.418 V$

All node voltages are determined. Now, the power of voltage sources can be calculated from the node voltages. For each source, we need to find the voltage across the source as well as the current flowing through it to compute the power.

$2A$ current source:
The voltage across the $2A$ current source is equal to $V_2$ . However, the comply with the passive voltage convention, the current should be entering from the positive terminal of the defined voltage as shown below. Therefore, $V_{2A}=-V_2=2.127 V$ .

$P_{2A}=2A \times V_{2A}=4.254W$ absorbing power

$1A$ current source:

To compliant with the passive sign convention, the voltage $V_{1A}$ should be defined with polarity as indicated above. We have $V_{1A}= V_5-V_4=-1.806V$ . Hence,

$P_{1A}=1A \times (-1.806 V)=-1.806W$ supplying power.

$5V$ voltage source:

$I_{5V}$ should be defined such that it enters from the positive terminal of the source in order to use the voltage of the source in power calculation. Another option is to use $V_3$ and define the current as entering from the voltage source terminal connected to the node of $V_3$ . We use the first approach here. KCL should be applied in the node of $V_3$ to determine $I_{5V}$ .

KCL @ Node of $V_3$ :

$-I_{5V}+I_{1\Omega}+I_{5\Omega}=0$
$\to I_{5V}=\frac{V_3-V_5}{1\Omega}+\frac{V_3-V_1}{5\Omega}=-0.7104A$

$P_{5V}=5V \times (-0.7104 A)= -3.552 W$ supplying power.

$3V$ voltage source:

Likewise, $I_{3V}$ should be defined as shown above to comply with the passive sign convention. We apply KCL to the reference node to find $I_{3V}$ .

KCL @ the reference node:

$I_{5V}+I_{3V}+2A=0$
$\to I_{3V}=-1.29A$

$P_{3V}=3V \times (-1.29 A)= -3.87 W$ supplying power.

The dependent source:
The voltage of the dependent source is $I_x$ and we define its current $I_{I_x}$ with the direction illustrated above. $I_{I_x}$ can be calculated by applying KCL at the node of $V_2$ . The current of the $3\Omega$ resistor is $I_x$ which is equal to $\frac{V_2}{3} +1= 0.291A$ .

KCL @ Node of $V_2$ :

$-2A+I_x+I_{I_x}=0 \to I_{I_x}=1.709A$

$P_{I_x}=I_x \times I_{I_x}=0.291 V \times 1.709 A= 0.497W$ absorbing power.

The PSpice simulation result is shown below. The PSpice schematics can be downloaded from http://www.solved-problems.com/download/1-23.zip.

Nodal Analysis – 6-Node Circuit

admin — Thu, 01 Jul 2010 05:44:17 +0000

Determine the power of each source after solving the circuit by the nodal analysis.

Solution
I. Identify all nodes in the circuit.
The circuit has 6 nodes as indicated below.

II. Select a reference node. Label it with the reference (ground) symbol.

The bottom left node is connected to 4 nodes while the other ones are connected to three or less elements. Therefore, we select it as the reference node of the circuit.

III. Assign a variable for each node whose voltage is unknown.
We label the remaining nodes as shown above. $V_1$ is connected to the reference node through a voltage source. Therefore, it is equal to the voltage of the voltage source: $V_1=10V$ .

IV. If there are dependent sources in the circuit, write down equations that express their values in terms of node voltages.

There is no dependent voltage source here.

V. Write down a KCL equation for each node.

Nodes of $V_2$ and $V_3$ are connected by a voltage source. Therefore, they form a supernode. The negative terminal of the voltage source is connected to $V_3$ and the positive terminal is connected to $V_2$ . Thus, $V_2=V_3+2.$ This can also be verified by a KVL around the loop which starts from the reference node, jumps to the node of $V_3$ with $-V_3$ (the reference is always assumed to be the negative terminal of node voltages), passes through the voltage source by $-2V$ and returns back to the reference node from $V_2$ as $+V_2$ $-V_3-2V+V_2=0 \to V_2=V_3+2.$
Supernode of $V_2$ & $V_3$ :

$\frac{V_3}{1\Omega}+\frac{V_3-V_4}{5\Omega}+\frac{V_2-V_1}{2\Omega}+\frac{V_2-V_5}{3\Omega}=0$
$\to {V_3}+\frac{V_3}{5}-\frac{V_4}{5}+\frac{V_3+2-10}{2}+\frac{V_3+2-V_5}{3}=0$
$\to {V_3}+\frac{V_3}{5}-\frac{V_4}{5}+\frac{V_3}{2}-4+\frac{V_3}{3}+\frac{2}{3}-\frac{V_5}{3}=0$
$\to \frac{61}{30}V_3-\frac{V_4}{5}-\frac{V_5}{3}=\frac{10}{3}$
$\to 61 V_3-6 V_4-10 V_5=100$

Node of $V_4$ :
$\frac{V_4}{6\Omega}+\frac{V_4-V_3}{5\Omega}+10=0$
$\to 5 V_4+6 V_4-6V_3+300=0$
$\to -6V_3+11 V_4=-300$

Node of $V_5$ :
$\frac{V_5}{4\Omega}+\frac{V_5-V_2}{3\Omega}-10=0$
$\to 3V_5+4V_5-4V_2-120=0$
$\to 7V_5-4 V_3-8-120=0$
$\to -4 V_3+7V_5=128$

Hence, we have the following system of equations:

$\left\{ \begin{array}{l} 61 V_3-6 V_4-10 V_5=100 \\ -6V_3+11 V_4=-300 \\ -4 V_3+7V_5=128 \end{array} \right.$

This system of equations can be solved by any preferred method such as elimination, row reduction, Cramer's rule or other methods. We use the Cramer's rule here:

$V_3=\frac{ \left| \begin{array}{c c c} 100 & -6 & -10 \\ -300 & 11 & 0 \\ 128 & 0 & 7 \end{array} \right| }{ \left| \begin{array}{c c c} 61 & -6 & -10 \\ -6 & 11 & 0 \\ -4 & 0 & 7 \end{array} \right| }=\frac{9180}{4005}=2.292 V$

$V_4=\frac{ \left| \begin{array}{c c c} 61 & 100 & -10 \\ -6 & -300 & 0 \\ -4 & 128 & 7 \end{array} \right| }{ \left| \begin{array}{c c c} 61 & -6 & -10 \\ -6 & 11 & 0 \\ -4 & 0 & 7 \end{array} \right| }=\frac{-104220}{4005}=-26.022 V$

and

$V_5=\frac{ \left| \begin{array}{c c c} 61 & -6 & 100 \\ -6 & 11 & -300 \\ -4 & 0 & 128 \end{array} \right| }{ \left| \begin{array}{c c c} 61 & -6 & -10 \\ -6 & 11 & 0 \\ -4 & 0 & 7 \end{array} \right| }=\frac{78480}{4005}=19.595 V.$

Thus,
$V_2=V_3+2=4.292.$

All node voltages are found. The current of the $10 V$ source is the current of the $2\Omega$ resistor, which is $\frac{V_2-V_1}{2\Omega}=-2.854A$ The current direction shosen such that the current enters from the positive terminal of the voltage source. This is only to comply with the passive sign convention. Now that we have the source current, its power can be easily calculated:

$P_{10V}=10 \times (-2.854)=-28.54 W$ absorbing power

The current of the $2V$ source equals to the summation of the currents of $5\Omega$ and $1\Omega$ resistors. Therefore,
$I_{2V}=I_{5\Omega}+I_{1\Omega}=\frac{V_3-V_4}{5\Omega}+\frac{V_3}{1\Omega}=5.663+2.292 =7.955.$
Consequently,

$P_{2V}=2 \times 7.955=15.91W$ supplying power.

The voltage across the $10A$ current source is $V_4-V_5=-45.617$ . Therefore,
$P_{10 A}=10 \times (-45.617)=-456.17 W$ supplying power.

The PSpice simulation result is indicated below. The PSpice schematics can be downloaded from http://www.solved-problems.com/download/1-22.zip.

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