Solved Problems

Category: Electrical Circuits Problems

All Solved Problems in Electrical Circuits

Nodal Analysis – Dependent Current Source

Deploy nodal analysis method to solve the circuit and find the power of the dependent source.

I. Identify all nodes in the circuit.
The circuit has 4 nodes.
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December 29, 2024
Nodal Analysis – Dependent Voltage Source

Use nodal analysis method to solve the circuit and find the power of the $3\Omega$ – resistor.

I. Identify all nodes in the circuit.
The circuit has 3 nodes:
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December 29, 2024
Nodal Analysis – Circuit with Dependent Voltage Source

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 this node 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 variables for unknown node voltages.
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. Incorporate dependent sources. 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. Apply Kirchhoff’s Current Law (KCL) / VI. Handle super-nodes.
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 $V_5$ :

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

VII. Solve the System of Equations.

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 $V_1=\frac{2}{3} V_2-1$ ,
$V_1=-2.418 V$

VIII. Determine Additional Variables.

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.

Download the Circuit File

To download the LTspice circuit file for your own simulations, click the link below.Please remember to unzip the file after downloading to access the .asc file for your simulations:

Nodal Analysis – Dependent Voltage Source and Super-node Circuit.asc Download

December 29, 2024
Nodal Analysis – 6-Node Circuit
Determine the power of each source after solving the circuit by the nodal analysis.

I. Identify all nodes in the circuit.

The circuit has 6 nodes as indicated below:

II. Select a reference node. Label this node 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 variables for unknown node voltages.

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. Incorporate dependent sources. There are no dependent sources in this circuit.

V. Apply Kirchhoff’s Current Law (KCL) / VI. Handle super-nodes.

Nodes of $V_2$ and $V_3$ are connected by a voltage source. Therefore, they form a super-node. 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 $+V_2$

$-V_3-2V+V_2=0 \to V_2=V_3+2.$

Super-node 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:

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

VII. Solve the System of Equations.

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:

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

$\text{[math]}$

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

$and$

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

$Thus, [latex]V_2=V_3+2=4.292.$

VIII. Determine Additional Variables.

After solving for all node voltages, we can calculate the currents and powers for the sources in the circuit as follows:
- Current of the $10 \, \mathrm{V}$ Source:
The current through the $10 \, \mathrm{V}$ source is the same as the current through the $2 \, \Omega$ resistor. Using Ohm’s Law, this current is:

$I_{10 \, \mathrm{V}} = \frac{V_2 - V_1}{2 \, \Omega} = -2.854 \, \mathrm{A}.$

The current direction is chosen such that it enters the positive terminal of the voltage source, in compliance with the \textit{passive sign convention}.

The power delivered by the $10 \, \mathrm{V}$ source is then:

$P_{10 \, \mathrm{V}} = 10 \times (-2.854) = -28.54 \, \mathrm{W}.$

Since the power is negative, the source is \textbf{absorbing power}.
- Current of the $2 \, \mathrm{V}$ Source:
The current through the $2 \, \mathrm{V}$ source is the sum of the currents through the $5 \, \Omega$ and $1 \, \Omega$ resistors. Using Ohm’s Law, we calculate:

$I_{2 \, \mathrm{V}} = I_{5 \, \Omega} + I_{1 \, \Omega} = \frac{V_3 - V_4}{5 \, \Omega} + \frac{V_3}{1 \, \Omega}.$

Substituting the values:

$I_{2 \, \mathrm{V}} = 5.663 + 2.292 = 7.955 \, \mathrm{A}.$

The power delivered by the $2 \, \mathrm{V}$ source is:

$P_{2 \, \mathrm{V}} = 2 \times 7.955 = 15.91 \, \mathrm{W}.$

Since the power is positive, the source is supplying power.
- Voltage Across the $10 \, \mathrm{A}$ Current Source:
The voltage across the $10 \, \mathrm{A}$ current source is given by the difference in node voltages:

$V_{10 \, \mathrm{A}} = V_4 - V_5 = -45.617 \, \mathrm{V}.$

The power delivered by the $10 \, \mathrm{A}$ source is:

$P_{10 \, \mathrm{A}} = 10 \times (-45.617) = -456.17 \, \mathrm{W}.$

Since the power is negative, the source is \textbf{supplying power}.

Download the Circuit File

To download the LTspice circuit file for your own simulations, click the link below.Please remember to unzip the file after downloading to access the .asc file for your simulations:

Nodal Analysis – 6-Node.asc Download
December 29, 2024
Nodal Analysis – Five-Node Circuit

Use nodal analysis to solve the circuit shown below and determine the power of node $V_{s_2}$ .

I. Identify all nodes in the circuit.
The circuit has five nodes as shown below.
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December 25, 2024
Solving a Simple Circuit of Three Elements

Solved the circuit to determine $I_x$ and power absorbed or supplied by each element.

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October 22, 2019
Power and Energy Conservation

Is the source Vs in the circuit absorbing or supplying power, and how much?

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August 28, 2019
Circuit Containing Only Sources

We go through solving a circuit which only containes independent sources: two voltage sources and two current sources. KVL and KCL are used to determine voltages and currents.

Determine the amount of power absorbed or supplied by each source.

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August 28, 2019
Find the current using superposition.

Find the current passing through the 2Ω resistor using superposition.

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August 12, 2019
AC Circuit Analysis – Sources with Different Frequencies

In AC circuit analysis, if the circuit has sources operating at different frequencies, Superposition theorem can be used to solve the circuit. Please note that AC circuits are linear and that is why Superposition theorem is valid to solve them.

Problem

Determine $i_x(t)$ where $i_s(t)=4 \cos (4t) ~ A$ and $v_s(t)=2 \sin (2t) ~ V$ .

Solution with AC Circuit Analysis

Since sources are operating at different frequencies, i.e. $4 \frac{rad}{s}$ and $2 \frac{rad}{s}$ , we have to use the Superposition theorem. That is to say that we need to determine contribution of each source on $i_x(t)$ . Then, the final answer is to obtained by adding the individual responses in the time domain. Please note that, since the impedances depend on frequency, we need to have a different frequency-domain circuit for each frequency.

Contribution of the current source

To find the contribution of the current source, we need to turn off other source(s). So, we need to turn off the voltage source. This is very similar to DC circuits that we discussed before:

Voltage sources become a short circuit when turned off.

Frequency domain

We first convert the circuit to the frequency domain:
$i_s(t)=4 \cos (4t) ~ A \rightarrow I_s=4\angle 0^{\circ} (\omega = 4 \frac{rad}{s})$
$L=1\text{H} \rightarrow Z_L=j \omega L=j4$
$C=1\text{F} \rightarrow Z_C=\frac{1}{j \omega C}=\frac{1}{j4}=-j0.25$
$R=1 \Omega \rightarrow Z_R=R=1$
$i_{x_1}(t) \rightarrow I_{x_1}$

Using current divider:
$I_{x_1}=\frac{Z_C}{Z_C+R}I_s$
$=\frac{-j0.25}{-j0.25+1} \times 4$
$=\frac{-j0.25}{1-j0.25} \times 4$
$=\frac{-j0.25 \times (1+j0.25)}{(1-j0.25)\times (1+j0.25)} \times 4$
$=\frac{-j0.25 + 0.0625)}{1^2+0.25^2} \times 4$
$=\frac{-j0.25 + 0.0625)}{1.0625} \times 4$
$=0.235-j0.941$
$=0.97\angle -76^{\circ}$

Time domain

Conversion to time-domain:
$i_{x_1}(t)=0.97 \cos (4t-76^{\circ}) ~ \text{A}$
Please note that the sinusoidal function for $i_s(t)$ is cosine and consequently, cosine must be used in converting $I_{x_1}$ to the time domain.

Contribution of the voltage source

To find the contribution of the voltage source, the current source needs to be turned off. As mentioned before:

To turn off a current source it should be replaced by an open circuit

Frequency domain

$v_s(t)=2 \sin (2t) ~ V \rightarrow V_s=2\angle 0^{\circ} (\omega = 2 \frac{rad}{s})$
$L=1\text{H} \rightarrow Z_L=j \omega L=j2$
$C=1\text{F} \rightarrow Z_C=\frac{1}{j \omega C}=\frac{1}{j2}=-j0.5$
$R=1 \Omega \rightarrow Z_R=R=1$
$i_{x_2}(t) \rightarrow I_{x_2}$

As the inductor branch is open, this is a very simple circuit with three elements in series: $R$ , $Z_C$ and $V_s$ . Therefore,
$I_{x_2}=\frac{V_s}{R+Z_C}$
$=\frac{2}{1-j0.5}$
$=\frac{2\times (1+j0.5)}{(1-j0.5)\times (1+j0.5)}$
$=\frac{2+j}{1^2+0.5^2}$
$=\frac{2+j}{1.25}$
$=1.6+j0.8$
$=1.789\angle 26.6^{\circ}$

Time domain

$i_{x_2}(t)=1.789 \sin (2t+26.6^{\circ}) ~ \text{A}$
(why $\sin$ ?)

Answer

Now that we have determined both $i_{x_1}(t)$ and $i_{x_2}(t)$ in time domain, we can go ahead and add them up to find $i_x(t)$ . Please note that we could not add $I_{x_1}$ and $I_{x_2}$ because they are not phasors with the same frequency.
$i_{x}(t)=i_{x_1}(t)+i_{x_2}(t)=0.97 \cos (4t-76^{\circ}) + 1.789 \sin (2t+26.6^{\circ}) ~ \text{A}$

April 23, 2015