Electrical Circuits Articles

Turning Sources Off

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.

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:

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:

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:

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.

Reference Node and Node Voltages

Reference Node
In circuits, we usually label a node as the reference node also called ground and define the other node voltages with respect to this point. The reference node has a potential of $0 V$ by definition. The following symbol is used to indicate the reference node:

The Reference Node Symbol

As mentioned, the selection of the reference node is arbitrary. However, a wise selection can make the solving easier. As a general rule, it 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.

Node Voltages
The voltage drop from a node to the reference node (ground) is called the node voltage. To keep definition simple, node voltages are usually defined with positive polarities.
Let’s find label node voltages in the following circuit:
A 5-Node Resistive Circuit

The circuit has 5 nodes:

Two of the nodes have 4 elements connected to them. These are the best candidates to be reference point.
Candidates for reference nodes

Let’s label one as reference node.
Label the reference node

Now, we define node voltages for the remaining nodes. These node voltages represent the voltage between the node and the reference.
Labeling node voltages

When there is a voltage source between a node and the reference node, the node voltage corresponds exactly to the voltage of the voltage source. In our example, we have two node voltages. The $-5 V$ voltage source is placed between the reference and the node labeled as $V_1$ . Therefore, $V_1=-5 V$ .
If there is a voltage source between two nodes, the difference between the corresponding node voltages equals to the voltage of the source. In our example, the $10 V$ voltage source is located between nodes labeled by $V_2$ and $V_4$ . Therefore, $V_2 - V_4= 10 V$ . It is important to note that voltage of the positive node minus the one of negative node is equal to the voltage of the source. KVL can be used to show this:
Voltage source between tow nodes

KVL around the loop: $-V_2+(+10)+V_4=0 to V_2-V_4=10 V$ . Recall that the reference node is always defined to be the negative polarity of all node voltages.

Voltage Divider – Voltage Division Rule

The voltage division rule (voltage divider) is a simple rule which can be used in solving circuits to simplify the solution. Applying the voltage division rule can also solve simple circuits thoroughly. The statement of the rule is simple:

Voltage Division Rule: The voltage is divided between two series resistors in direct proportion to their resistance.

It is easy to prove this. In the following circuit

Voltage Divider

the Ohm’s law implies that
$v_1(t)=R_1 i(t)$ (I)
$v_2(t)=R_2 i(t)$ (II)
Applying KVL
$-v(t)+v_1(t)+v_2(t)=0 \rightarrow v(t)=v_1(t)+v_2(t)$ .

Therefore
$v(t) = R_1 i(t)+R_2 i(t)= (R_1 +R_2) i(t)$ .
Hence
$i(t) = \frac{v(t)}{R_1 +R_2}$ .
Substituting in I and II
$v_1(t)=R_1 \frac{v(t)}{R_1 +R_2}$ ,
$v_2(t)=R_2 \frac{v(t)}{R_1 +R_2}$ .
Consequently

$v_1(t)= \frac{R_1}{R_1 +R_2} v(t)$ ,
$v_2(t)=\frac{R_2}{R_1 +R_2} v(t)$ .

which shows that the voltage is divided between two series resistors in direct proportion to their resistance. The rule can be easily extended to circuits with more than two resistors. For example,

Voltage Division among four resistors

$v_1(t)= \frac{R_1}{R_1 +R_2+R_3+R_4} v(t)$ ,
$v_2(t)=\frac{R_2}{R_1 +R_2+R_3+R_4} v(t)$ ,
$v_3(t)=\frac{R_3}{R_1 +R_2+R_3+R_4} v(t)$ ,
$v_4(t)=\frac{R_4}{R_1 +R_2+R_3+R_4} v(t)$ .
The voltage division rule can be used solve simple circuits or to simplify solving complicated circuits.

Problem 1-16: Voltage Divider

Find $V_x$ (or $v_x(t)$ ) and $I_x$ (or $i_x$ ) using voltage division rule.
a)
Voltage Divider Problem - A

Solution
a)
Voltage Divider Problem - A

Voltage divider: $V_x=\frac{5\Omega}{2\Omega+5\Omega}\times 14 V=10 V$
Ohm’s law: $I_x=\frac{V_x}{5 \Omega}=2 A$
b)
Voltage Divider Problem - B

Voltage divider: $V_x=\frac{4\Omega}{2\Omega+3\Omega+1\Omega+4\Omega}\times (-9 V)=-3.6 V$
Ohm’s law: $I_x=-\frac{V_x}{4 \Omega}=0.9 A$
Please note that $I_x$ is leaving from the positive terminal of $V_x$ . Therefore, applying the Ohm’s law results in $V_x=-R\times I_x$ .
c)
Voltage Divider Problem - C

Voltage divider: $v_x(t)=\frac{5\Omega}{2\Omega+5\Omega+3\Omega}\times (-5 \sin (2t))=-2.5 \sin (2t) V$
Ohm’s law: $i_x(t)=\frac{v_x(t)}{5 \Omega}=-\frac{1}{2} \sin (2t) A$
d)
Voltage Divider Problem - D

The tricky part in this problem is the polarity of $V_x$ . In the defined formula for voltage divider, the current is leaving the voltage source from the positive terminal and entering to resistors from positive terminals. In this problem, the current is entering to the the resistor from the negative terminal. Therefore, the voltage for $V_x$ is the negative of the voltage obtained from the voltage divider formula. The reason is that another voltage can be defined with the inverse polarity and its value can be found using the voltage division rule. $V_x$ is the negative of the defined voltage because it represents the voltage across the same nodes with inverse polarity.
Voltage divider: $V_x=- \frac{5\Omega}{2\Omega+5\Omega}\times 10 V= - 10 V$
Ohm’s law ( $I_x$ is entering from the negative terminal of $V_x$ ): $I_x= - \frac{V_x}{5 \Omega}=2 A$
.

One of the common mistakes in using the voltage division rule is to use the formula for resistors which are in parallel with other elements. For example, the voltage division rule cannot be used in the following circuit directly.
voltage divider - misleading case - a

It will be incorrect if one tries to find $V_x$ using voltage divider by neglecting the other $6 \Omega$ resistor as
voltage divider - misleading case - b

So, $V_x \neq \frac{6 \Omega}{2 \Omega + 6 \Omega} 15 V$ . However, if solving other parts of a circuits confirms that the current of the other element/branch is zero, the voltage division rule can be still applied. For example, suppose that the following network is a piece of a larger circuit.
voltage divider - piece of a larger circuit

voltage divider - piece of a larger circuit

Let’s assume that the analysis of the circuit shows that $I_x=0$ . In this case, $V_x= \frac{6 \Omega}{2 \Omega + 6 \Omega} 12 V = 9 V$ regardless of where A and B are connected.

Ideal Independent Sources

1) Ideal Independent Voltage Sources
An ideal independent voltage source is a two-terminal circuit element where the voltage across it
a) is independent of the current through it
b) can be specified independently of any other variable in a circuit.
There are two symbols for ideal independent voltage source in circuit theory:

Symbol for Constant Independent Voltage Source

General Symbol for Independent Voltage Source

The battery cell symbol is usually used for constant voltage sources but the other one is more general and can be used for both variable and constant voltage sources.
The v-i plot for an ideal independent voltage source is shown below.

v-i plot for ideal independent voltage source

In a circuit, voltage across elements which are parallel with voltage sources are equal to the voltage of the corresponding voltage sources. This is a useful rule in solving circuits. For example, check out the following problem.

Problem 1-13: Voltage of A Current Source

Find voltages across the current sources.
a)
A Current Source Parallel with A Voltage Source

Parallel Current and Variable Voltage Sources

Solution

In each case, the current source is parallel with a voltage source. Therefore, the voltage across the current source is equal to the voltage of the voltage source, regardless of other elements.
a)
A Current Source Parallel with A Voltage Source

2) Ideal Independent Current Sources
In contrast to ideal independent voltage sources, an ideal independent current source is a two-terminal circuit element where the current passing through it
a) is independent of the voltage across it
b) can be specified independently of any other variable in a circuit.
There is one symbol for ideal independent current source in circuit theory:

Symbol for Ideal Independent Current Source

The v-i plot for an ideal independent current source is depicted below.

v-i plot for ideal independent current source

Similar to voltage sources, the charactristic of ideal independent current sources can also be used to solve circuits (or a portion of a circuit) where an element is in series with a current source. Lets consider the following solved problem.

Problem 1-14: Current of A Voltage Source

Find the current passing through the voltage source:
a)
A voltage source in series with a current source

A voltage source in series with a current source and other elements

Solution
a) The voltage source is in series with the current source. Since by definition a current source keeps the current passing through itself constant and the voltage source is in series with the current source, it should have the same current $10 A$ .
A voltage source in series with a current source

b) The same scenario is happening here and the voltage source has the same current as the current source $5A$ .
A voltage source in series with a current source and other elements

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Tayyab Siddiqui