Introduction to Electrical and Electronics Engineering. Module 1, module 1.

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Important note

• If current is not given in a circuit then voltage division rule can use
• If voltage not given in a circuit then current division rule can use
• If voltage and current available in a circuit use ohms law
• Specifically asked in the question apply the rules

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Find voltage drop across each resistance.

Applying the voltage division rule voltage across $200\,\Omega$ resistance we get

$E_1 = \frac{E}{r_1+r_2} r_1$

$E_1 = \frac{6}{200+100} \text{ X } 200 = 4\text{V}$

Voltage across $100\,\Omega$ resistance we get

$E_2 = \frac{E}{r_1+r_2} r_2$

$E_2 = \frac{6}{200+100} \text{ X } 100 = 2\text{V}$

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Consider two resistors $20\,\Omega$ and $40\,\Omega$ are connected in a parallel with a current source of $20\text{ A}$. Find out the current flowing through each resistor in the parallel circuit.

Applying the current division rule voltage across $20\,\Omega$ resistance we get

$I_1 = \frac{I}{r_1+r_2} r_2$

$I_1 = \frac{20}{20+40} \times 40 = 13.33\text{A}$

Voltage across $40\,\Omega$ resistance we get

$I_2 = \frac{I}{r_1+r_2} r_1$

$I_2 = \frac{20}{20+40} \times 20 = 6.67\text{A}$

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Example 1.11.1 A resistor of $5\;\Omega$ is connected in series with a parallel combination of $6\;\Omega$ and $3\;\Omega$. Find the supply current taken from a $35\;\text{V}$ dc source.

Solution

Let $R_P$ be the equivalent resistance of the parallel combination of $3\;\Omega$ and $6\;\Omega$.

Then $\dfrac{1}{R_P} = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3}{6}$

Therefore, $R_P = 2\;\Omega$

Total resistance in the circuit, $R = R_P + 5 = 2 + 5 = 7\;\Omega$

Supply current, $I = \dfrac{V}{R} = \dfrac{35}{7} = 5\;\text{A}$

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Example 1.11.2 Find the total resistance of the circuit across the terminals X and Y. Also find the power consumed by the circuit if a $12\text{ V}$ battery is connected across XY.

Figure 1.13

Solution

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$R_{XZ} + R_{ZY} = 1 + 1 = 2\, \Omega$ $\frac{1}{R_{XY}} = \frac{1}{4} + \frac{1}{2} + \frac{1}{4} = 1$

Therefore, equivalent resistance, $R_{XY} = 1\, \Omega$

Total current in the circuit, $I = \frac{V}{R_{XY}} = \frac{12}{1} = 12\text{ A}$

Power consumed $= VI = 12 \times 12 = 144\text{ W}$

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Example 1.11.3 Three resistors $10\ \Omega$, $20\ \Omega$ and $30\ \Omega$ are connected in series across a $100\text{ V}$ supply. Find the voltage across each resistor.

Solution

Figure 1.15

Let $R_S$ be the equivalent series resistance

$R_S = 10 + 20 + 30 = 60\ \Omega$

$I = \frac{V}{R_S} = \frac{100}{60} = 1.666\text{ A}$

Voltage drop across $10\ \Omega$, $V_1 = 1.666 \times 10 = 16.66\text{ V}$

Voltage drop across $20\ \Omega$, $V_2 = 1.666 \times 20 = 33.33\text{ V}$

Voltage drop across $30\ \Omega$, $V_3 = 1.666 \times 30 = 50\text{ V}$

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Example 2.6.2 Find the voltage drop and current through all the resistances in the circuit shown in Figure 2.20.

Figure 2.20

Solution

$\text{Equivalent resistance} = 6 + \frac{12 \times 6}{12 + 6} = 10\;\Omega$

$\text{Source current}, I = \frac{\text{Source voltage}}{\text{Equivalent resistance}} = \frac{50\;\text{V}}{10\;\Omega} = 5\;\text{A}$

Voltage drop across $6\;\Omega$ resistor

$= \text{Current through } 6\;\Omega \times \text{Resistance}$

$= 5 \times 6 = 30\;\text{V}$

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Using current divider rule, current through $12\ \Omega$ resistor $= I_1$

$\quad = 5 \times \frac{6}{12 + 6} = \frac{5}{3}\text{ A}$

By Kirchhoff's current law, $I = I_1 + I_2$

Therefore, $I_2 = I - I_1 = 5 - \frac{5}{3} = \frac{10}{3}\text{ A}$

Voltage drop across $12\ \Omega = I_1 \times 12 = \frac{5}{3} \times 12 = 20\text{ V}$

Voltage drop across $6\ \Omega = I_2 \times 6 = \frac{10}{3} \times 6 = 20\text{ V}$

The voltage across $6\ \Omega$ and $12\ \Omega$ resistors are the same because they are connected in parallel.

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Application 2

• Ohms law

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Source transformation rule

Voltage source to current source

Figure 2.5: Voltage source to current source transformation

A voltage source $V_s$ in series with resistance $R$ can be transformed into a current source $I_s$ in parallel with resistance $R$ where $I_s = \frac{V_s}{R}$

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Current source to voltage source

Figure 2.6: Current source to voltage source transformation

Similarly, a current source $I_S$ in parallel with a resistance $R$ can be transformed into a voltage $V_S$ series with a resistance $R$ where $V_S = I_S R$

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• Node: A node is a junction where two or more network elements are connected together.

• Branch: An element or number of elements connected between two nodes constitute a branch.

• Junction: A point where three or more circuit components meet.

• Electrical network: An electrical network is an interconnection of electrical elements.

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Important points

• NODE
• JUNCTION
• BRANCH
• LOOP
• MESH
• Electric network

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Circuit vs Network

• Electrical circuit: Electric circuit is an electrical network with at least one closed path.
• All circuits are networks; but, all networks are not circuits.
• Loop: A loop is any closed part of the circuit.
• Mesh: A mesh is the most elementary form of a loop and cannot be further divided into other loops.
• All meshes are loops but all loops are not meshes

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1. Absolute (Node) Potential ($V_A$): The voltage at Node A measured relative to the reference node (Ground, which is assumed to be $0\text{ V}$).

2. Relative Potential ($V_{AB}$): The voltage at Node A relative to Node B.

The Formula:

$V_{AB} = V_A - V_B$

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Relative potential

• Relative potential (or potential difference) is the voltage of one point in a circuit measured with respect to another point, rather than to ground.

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A branch is a single path containing a circuit component connected between two nodes

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Finding Loops in a Circuit

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Mesh

• A closed loop that doesn't contain any other loop within it is referred to as a mesh. In other words, any path within a circuit that doesn't include any other path is considered a mesh

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The overall circuit depicted in fig-1 comprises 6 Nodes, 7 Branches, 3 Loops, and 2 Meshes, as illustrated in fig-6.

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Robert Kirchhoff was a German physicist and mathematician who contributed to the fundamental understanding of electrical circuits

• Kirchhoff (1824–1887)

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KIRCHHOFF'S VOLTAGE LAW

• Kirchhoff's voltage law states that the algebraic sum of all voltages in a closed path in a circuit is always zero at all instants of time.
• (Or) sum of potential drop is equal to sum of potential rises

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Polarities

Potential rise

Potential drops

$V + V_1 + V_2 + V_3 + V_4 = 0$

V across the resistance $R_1 = IR_1$

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Potential rise

Potential drop

Current direction important(i1-i2) next loop (i2-i1) same direction Current direction important(i1+i2) next loop (i2+i1) opp direction

Voltage polarity very important Always start with voltage source for writing equations

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Sign convention for applying signs to the voltage polarities in our KVL equations

• the loop, if the positive terminal is encountered before the negative terminal, the voltage is interpreted as positive in the KVL equation.
• If the negative terminal is encountered first, the voltage is interpreted as negative in the KVL equation.

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Example kvl equation

If we travel in clock-wise direction,

$+V - IR_1 - IR_2 - IR_3 - IR_4 = 0$

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1 mesh problems

KVL BASED PROBLEM

Example 2.6.1 Find the current in the circuit shown in Figure 2.

Solution

Let the current $I$ be flowing in clockwise direction. Marking the polarities of voltages across resistors as per sign convention as shown in Figure 2.19.

Figure 2.19

Applying KVL, we get,

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KVL BASED PROBLEM

Example 2.6.1 Find the current in the circuit shown in Figure 2.18.

Figure 2.18

Solution

Let the current $I$ be flowing in clockwise direction. Marking the polarities of voltages across resistors as per sign convention as shown in Figure 2.19.

Figure 2.19

Applying KVL, we get,

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Example 2.6.5 Find the current through $4 \; \Omega$ in the circuit shown in Figure 2.23.

NODE

2 mesh problems

KVL PROBLEM

all mesh currents are drawn clockwise by default, shared branches always have opposing currents,

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KVL PROBLEM

Solution

From the loop 1: $15 - 10I_1 - 5(I_1 - I_2) = 0$

$15I_1 - 5I_2 = 15$

$3I_1 - I_2 = 3 \quad (2.4)$

From the loop 2: $-5(I_2 - I_1) - 4I_2 - 9 = 0$

$9I_2 - 5I_1 = -9 \quad (2.5)$

From the expressions 2.4 and 2.5, we get, $I_1 = 0.82 \text{ A}$

$I_2 = -0.54 \text{ A}$

Note: The current flowing through the $4\,\Omega$ resistor is the mesh current $I_2$:

$I_{4\Omega} = I_2 = - \frac{6}{11} \text{ A} \approx -0.545 \text{ A}$

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Solve the System of Simultaneous Equations

From Equation 1, express $I_2$ in terms of $I_1$:

$I_2 = 3I_1 - 3$

Substitute $I_2$ into Equation 2回忆:

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Example 2.6.3 Find the loop currents in the electric circuit shown in Figure 2.21.

3 mesh problems

KVL PROBLEM

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Solution

From the loop 1 : $5 - 5I_1 - 1(I_1 - I_2) = 0$

6I_1 - I_2 = 5 \tag{2.1}

From the loop 2 : $-2I_2 - 2(I_2 - I_3) - 1(I_2 - I_1) = 0$

5I_2 - I_1 - 2I_3 = 0 \tag{2.2}

From loop 3 : $-2(I_3 - I_2) - 3I_3 + 10 = 0$

5I_3 - 2I_2 = 10 \tag{2.3}

Solving 2.1, 2.2 and 2.3, we get,

$I_1 = 1.03\text{ A}$

$I_2 = 1.198\text{ A}$

$I_3 = 2.48\text{ A}$

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From (Eq. 1): $I_2 = 6I_1 - 5$

From (Eq. 3): $5I_3 = 10 + 2I_2 = 10 + 2(6I_1 - 5) = 10 + 12I_1 - 10 = 12I_1 \implies I_3 = 2.4I_1$

Substitute $I_2$ and $I_3$ into (Eq. 2):

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Super mesh(note)

• when a current source is located on the boundary shared by two meshes.
• To solve this, we "remove" the shared current source branch and create a larger loop
• Series Resistors: If a resistor is in series with the shared current source, it is "removed" along with the source
• If a current source is on the outside of the circuit (not shared), you don't need a supermesh.

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points (Super mesh )

• A supermesh is created in circuit analysis when a current source (independent or dependent) is shared between two adjacent meshes.
• Temporarily remove the shared current source and any element connected strictly in series with it, leaving an open circuit.
• Apply Kirchhoff's Voltage Law (KVL)

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current source is located on the boundary shared by two meshes.

Example 1. Determine the current in the $5\text{ }\Omega$ resistor in the network given in Fig.

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$50 - 10(I_1 - I_2) - 5(I_1 - I_3) = 0$

$(1)$

$15I_1 - 10I_2 - 5I_3 = 50$

Step B: Write the Constraint Equation for the Supermesh

The $2\text{ A}$ current source is in a branch shared by Mesh 2 and Mesh 3.

• Current $I_2$ flows in the same direction as the source (right to left).
• Current $I_3$ flows in the opposite direction (left to right).

$I_2 - I_3 = 2 \implies I_2 = I_3 + 2 - (\text{Eq. 2})$

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$-2I_2 - I_3 - 5(I_3 - I_1) - 10(I_2 - I_1) = 0$ $\ -15I_1 + 12I_2 + 6I_3 = 0 \quad (3)$

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4. Solve the System of Equations

Substitute $I_2 = I_3 + 2$ from (Eq. 2) into (Eq. 1) and (Eq. 3):

From (Eq. 1):

$3I_1 - 2(I_3 + 2) - I_3 = 10 \implies 3I_1 - 3I_3 = 14 \implies I_1 - I_3 = \frac{14}{3}\text{ A}$

Interestingly, we don't need to go further because the question asks for the current in the $5\text{ }\Omega$ resistor. Looking at the diagram, the net downward current in the $5\text{ }\Omega$ resistor is exactly $(I_1 - I_3)$.

$I_{5\Omega} = I_1 - I_3 = \frac{14}{3}\text{ A} \approx 4.67\text{ A}$

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outside of the circuit (not shared), you don't need a supermesh.

Example 2. Write the mesh equations for the circuit shown in Fig. and determine the currents $I_1$, $I_2$ and $I_3$.

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Summing the voltage drops to zero:

$3(I_2 - I_1) + 10 + 2(I_2 - I_3) = 0$

Substitute $I_1 = 10$:

$3(I_2 - 10) + 10 + 2(I_2 - I_3) = 0$

$3I_2 - 30 + 10 + 2I_2 - 2I_3 = 0$

$5I_2 - 2I_3 = 20 \quad \text{--- (Equation 1)}$

$1 \cdot I_3 + 2(I_3 - I_2) - 10 = 0$

$-2I_2 + 3I_3 = 10 \quad \text{--- (Equation 2)}$

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3. Solving the system of equations

From equation 2,

$3I_3=2I_2+10,\qquad I_3=\frac{2I_2+10}{3}.$

Substitute into equation 1:

$5I_2-2\left(\frac{2I_2+10}{3}\right)=20.$

Multiplying by 3 and simplifying,

$15I_2-4I_2-20=60,\qquad 11I_2=80.$

$I_2=\frac{80}{11}\,\mathrm{A}\approx7.27\,\mathrm{A}.$

Then

$I_3=\frac{2(80/11)+10}{3}=\frac{270}{33}=\frac{90}{11}\,\mathrm{A}\approx8.18\,\mathrm{A}.$

Final currents

$I_1=10\,\mathrm{A},\qquad I_2=\frac{80}{11}\,\mathrm{A},\qquad I_3=\frac{90}{11}\,\mathrm{A}.$

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Mesh current and matrix Representation

2.6.1 Matrix method of analysis

An $n$-mesh network can be represented by $n$ simultaneous equations. Refer to the three-mesh network shown in Figure 2.25.

Figure 2.25

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The matrix equation for the network is

$R_{11} = 1^{\text{st}} \text{ loop total resistance}$

$R_{12} = \text{common resistance in between}$ $\text{loop } 1 \text{ and } 2$

$R_{12} = R_{21}$ $R_{23} = R_{32}$ $R_{13} = R_{31}$

$V_1 \text{ is } - \text{ to } + \text{ then } v_1$ $V_2 \text{ is } + \text{ to } - \text{ then } -v_2$

$\begin{bmatrix} R_{11} & -R_{12} & -R_{13} \\ -R_{21} & R_{22} & -R_{23} \\ -R_{31} & -R_{32} & R_{33} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}$

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Matrix form RULE

• Diagonal Elements ($R_{11}, R_{22}, R_{33}$): Always Positive ($+$)
  • Sum of all resistances in that specific loop.
• Off-Diagonal Elements ($R_{12}, R_{23}, R_{31}$, etc.): Usually Negative ($-$)
  • Sum of resistances shared between two loops.
  • Sign Rule: If both loop currents pass through the shared resistor in opposite directions, the entry is negative ($-$). If they flow in the same direction, the entry is positive ($+$). Since all mesh currents are drawn clockwise by default, shared branches always have opposing currents, making off-diagonal terms negative.

If two voltage source is same loop For summing voltage element same polarity must be minus the voltage Opposite polarity add the voltage

For voltage matrix voltage polarity is depends on current flow direction

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Applying KVL to each mesh, the following equations are obtained.

$\begin{bmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}$

Rearranging the above equations, we can write as

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These equations can be represented in matrix form as given below.

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Example 2.6.8 Determine the mesh currents in the network shown in Figure 2.26.
Figure 2.26.

Solution
The matrix equation for the network is
\begin{bmatrix}8&-3&0\\-3&9&2\\0&2&10\end{bmatrix}\begin{bmatrix}I_1\\I_2\\I_3\end{bmatrix}=\begin{bmatrix}5\\10\\35\end{bmatrix}\tag{2.19}

$\Delta_R = \begin{vmatrix} 8 & -3 & 0 \\ -3 & 9 & 2 \\ 0 & 2 & 10 \end{vmatrix} = 8(90 - 4) - 3(+30) = 598$

$\Delta_1 = \begin{vmatrix} 5 & -3 & 0 \\ 10 & 9 & 2 \\ 35 & 2 & 10 \end{vmatrix} = 5(90 - 4) + 3(100 - 70) = 520$

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$\Delta_2 = \begin{vmatrix} 8 & 5 & 0 \\ -3 & 10 & 2 \\ 0 & 35 & 10 \end{vmatrix} = 8(100 - 70) - 5(-30) = 390$

$\Delta_3 = \begin{vmatrix} 8 & -3 & 5 \\ -3 & 9 & 10 \\ 0 & 2 & 35 \end{vmatrix} = 8(315 - 20) + 3(-105) + 5(-6) = 2015$

$\therefore I_1 = \frac{\Delta_1}{\Delta_R} = \frac{520}{598} = 0.87\text{ A}$

$I_2 = \frac{\Delta_2}{\Delta_R} = \frac{390}{598} = 0.65\text{ A}$

$I_3 = \frac{\Delta_3}{\Delta_R} = \frac{2015}{598} = 3.37\text{ A}$

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Example 2.6.9 Write the resistance matrix of the network shown in Figure 2.27 and determine the current drawn from the voltage source.

Figure 2.27

Solution

The matrix equation for the network is $\begin{bmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}$

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\begin{bmatrix} 10 & -4 & -3 \\ -4 & 8 & -2 \\ -3 & -2 & 9 \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} 10 \\ 0 \\ 0 \end{bmatrix} \tag{2.20}

$\Delta_R = \begin{vmatrix} 10 & -4 & -3 \\ -4 & 8 & -2 \\ -3 & -2 & 9 \end{vmatrix} = 10(72 - 4) + 4(-36 - 6) - 3(8 + 24) = 416$

$\Delta_1 = \begin{vmatrix} 10 & -4 & -3 \\ 0 & 8 & -2 \\ 0 & -2 & 9 \end{vmatrix} = 10(72 - 4) + 4(0 - 0) - 3(0 - 0) = 680$

$\Delta_2 = \begin{vmatrix} 10 & 10 & -3 \\ -4 & 0 & -2 \\ -3 & 0 & 9 \end{vmatrix} = 10(0 - 0) - 10(-36 - 6) - 3(0 - 0) = 420$

$\Delta_3 = \begin{vmatrix} 10 & -4 & 10 \\ -4 & 8 & 0 \\ -3 & -2 & 0 \end{vmatrix} = 10(0 - 0) + 4(0 - 0) + 10(8 + 24) = 320$

$\therefore I_1 = \frac{\Delta_1}{\Delta_R} = \frac{680}{416} = 1.635\text{ A}$

$I_2 = \frac{\Delta_2}{\Delta_R} = \frac{420}{416} = 1.01\text{ A}$

$I_3 = \frac{\Delta_3}{\Delta_R} = \frac{320}{416} = 0.77\text{ A}$

Figure 2.27

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Home work

3. (a) Determine the current $I_L$ in the circuit shown in the given figure by mesh analysis using matrix form.

(10)

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KIRCHHOFF'S CURRENT LAW

• Kirchhoff's current law states any current flowing into a node must simultaneously equal the current flowing out of it at any given moment.
• If entering currents are assigned positive sign, leaving currents will be given negative sign.

$I_T = I_1 + I_2 + I_3$

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k.c.l Version 2

In the example below,

The algebraic sum of all currents meeting at a node is equal to zero.

A node is a junction where two or more network elements are connected together.

Algebraic Sum: $\sum I = 0$

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Find node voltages

• In this method, one of the independent nodes is taken as reference node ($0\text{v}$) and the voltages of the remaining independent nodes are assumed.

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• Consider the following circuit. The circuit has 3 independent nodes A, B and C. Let C be the reference node. Let VA and VB be the potential of A and B with respect to reference node • In node voltage analysis, to identify a node points, current must enter and leave

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Node voltage method step

• Find the node voltages so we can find the current flow through circuits and components

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There are two nodes A,B and 1 relative potential and 2 absolute potential differences

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Three branches connect to node A. Assuming all currents leave node A: A current is $I_1+I_2+I_3=0$

• Left Branch (towards $V_1$): Current flows from potential $V_A$ to potential $V_1$ across resistor $R_1$.

$I_{\text{left}} = \frac{V_A - V_1}{R_1}$

• Middle-Down Branch (towards Node C): Current flows from $V_A$ to ground ($0\text{ V}$) through $R_2$.

$I_{\text{down}} = \frac{V_A - 0}{R_2} = \frac{V_A}{R_2}$

• Right Branch (towards Node B): Current flows from $V_A$ to $V_B$ across $R_3$.

$I_{\text{right}} = \frac{V_A - V_B}{R_3}$

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Applying KCL at node A, $\rightarrow I_1 + I_2 + I_3 = 0 \rightarrow I = V/R$

$\frac{V_A - V_1}{R_1} + \frac{V_A}{R_2} + \frac{V_A - V_B}{R_3} = 0$

Applying KCL at node B,

$\frac{V_B - V_2}{R_5} + \frac{V_B}{R_4} + \frac{V_B - V_A}{R_3} = 0$

By solving the two equations, we can determine $V_A$ and $V_B$.

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Using nodal analysis, find current through $2\Omega$ resistance.

Example

First find the node voltages Then only we can find current

Ans:-

Applying KCL at node A, $\frac{V_A - 10}{6} + \frac{V_A}{1} + \frac{V_A - V_B}{2} = 0$

$V_A \left( \frac{1}{6} + 1 + \frac{1}{2} \right) - \frac{V_B}{2} = \frac{10}{6}$

$1.667 V_A - 0.5 V_B = 1.667 \quad ---\text{(1)}$

Applying KCL at node B, $\frac{V_B - 20}{10} + \frac{V_B}{3} + \frac{V_B - V_A}{2} = 0$

$-\frac{V_A}{2} + V_B \left( \frac{1}{10} + \frac{1}{3} + \frac{1}{2} \right) = \frac{20}{10}$

$-0.5 V_A + 0.933 V_B = 2 \quad ---\text{(2)}$

Solving the equations,

$V_A = 1.957\text{V}, \quad V_B = 3.192\text{V}$

$I = \frac{V_A - V_B}{2} = \frac{1.957 - 3.192}{2} = -0.617\text{A}$

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Example 2.7.1 Find current through the resistor $20\ \text{ }\ \text{\Omega}$ in the circuit shown in Figure 2.32, using nodal analysis.

Figure 2.32: Circuit for Example 2.7.1

First find the node voltages Then only we can find current

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e node $V_1$:

$\frac{V_1 - 8}{10} + \frac{V_1}{20} + \frac{V_1 - 2}{30} = 0$

Step 2: Solve for Node Voltage $V_1$

Multiply the entire equation by the common denominator $60$ to clear fractions:

$6(V_1 - 8) + 3V_1 + 2(V_1 - 2) = 0$

$6V_1 - 48 + 3V_1 + 2V_1 - 4 = 0$

$11V_1 - 52 = 0$

$11V_1 = 52$

Step 3: Calculate Current through the $20\ \text{}\Omega$ Resistor

The current flowing downward through the $20\ \text{}\Omega$ resistor is:

$I_{20\ \text{}\Omega} = \frac{V_1}{20} = \frac{4.727}{20} \approx 0.236\ \text{A} \quad \left(\text{or}\ \frac{13}{55}\ \text{A}\right)$

The current through the $20\ \text{}\Omega$ resistor is $0.236\ \text{A}$ (flowing downwards).

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Solution

Circuit is redrawn with node voltages labelled, resistors are designated and reference node as shown in Figure 2.33.

$\frac{V_b - 8}{10} + \frac{V_b - 0}{20} + \frac{V_b - 2}{30} = 0$

$V_b \left(\frac{1}{10} + \frac{1}{20} + \frac{1}{30}\right) - \left(\frac{8}{10} + \frac{2}{30}\right) = 0$

$\therefore V_b = 4.7\text{ V}$

$\text{Current through } 20\text{ }\Omega,\ i_2 = \frac{V_b}{20} = \frac{4.7}{20} = 0.24\text{ A}$

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Example 2.7.2 Find current through the resistor $R_1$ for the network shown in Figure 2.34.

Figure 2.34: Circuit for Example 2.7.2

First find the node voltages Then only we can find current

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$\frac{V_1 - 10}{10} + \frac{V_1}{5} - 1 = 0$

$V_b \left(\frac{1}{10} + \frac{1}{5}\right) = 2$

Solving for $V_b$, we get, $V_b = 6.67\text{ V}$

Current through resistor $R_1$, $i_1 = \frac{V_a - V_b}{R_1}\text{ A}$

$= \frac{10 - 6.67}{10} = 0.333\text{ A}$

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Example 2.7.4 Determine node voltages for the network shown in Figure 2.38.

Figure 2.38: Circuit for Example 2.7.4

First find the node voltages Then only we can find current

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Figure 2.39 and directi

From figure, $V_b = V_s = 100$

From figure, $V_b = 100$

$V_a \left(\frac{1}{10} - \frac{1}{20} + \frac{1}{30}\right) - V_c \left(\frac{1}{30}\right) = \frac{100}{20}$ $0.183V_a - 0.033V_c = 5$

$\frac{V_a - V_d}{R_1} + \frac{V_a - V_b}{R_2} + \frac{V_a - V_c}{R_3} = 0$

$\frac{V_a}{R_1} + \frac{V_a - 100}{R_2} + \frac{V_a - V_c}{R_3} = 0$

KCL at Node $V_c$: $\frac{V_c - V_a}{R_3} + \frac{V_c - V_b}{R_4} + \frac{V_c - V_d}{R_5} = 0$

Substituting $V_d = 0\text{ V}$ and $V_b = 100\text{ V}$ gives:

$\frac{V_c - V_a}{R_3} + \frac{V_c - 100}{R_4} + \frac{V_c}{R_5} = 0$

$V_a \left(\frac{1}{30}\right) - V_c \left(\frac{1}{30} + \frac{1}{40} + \frac{1}{50}\right) = - \frac{100}{40}$

$0.033V_a - 0.078V_c = -2.5$

Solving for $V_a$ and $V_c$, we get, $V_a = 35.84\text{ V}$ and $V_c = 47.21\text{ V}$

ideal voltage source cannot be expressed using Ohm's Law directly because an ideal voltage source has zero internal resistance.

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KCL at Node 1 ($V_1$): Sum of currents leaving Node 1 equals zero: $\frac{V_1 - 0}{10} + \frac{V_1 - 100}{20} + \frac{V_1 - V_2}{30} = 0$

Multiply the entire equation by $60$ to clear denominators: $6V_1 + 3(V_1 - 100) + 2(V_1 - V_2) = 0$ $6V_1 + 3V_1 - 300 + 2V_1 - 2V_2 = 0$ $11V_1 - 2V_2 = 300 \quad - (\text{Equation } 1)$

KCL at Node 2 ($V_2$): Sum of currents leaving Node 2 equals zero: $\frac{V_2 - 0}{50} + \frac{V_2 - 100}{40} + \frac{V_2 - V_1}{30} = 0$

Multiply the entire equation by $600$ to clear denominators: $12V_2 + 15(V_2 - 100) + 20(V_2 - V_1) = 0$ $12V_2 + 15V_2 - 1500 + 20V_2 - 20V_1 = 0$ $-20V_1 + 47V_2 = 1500 \quad - (\text{Equation } 2)$

Step 3: Solve the System of Equations From Equation 1, express $V_2$ in terms of $V_1$: $2V_2 = 11V_1 - 300 \implies V_2 = 5.5V_1 - 150$

Substitute $V_2$ into Equation 2: $-20V_1 + 47(5.5V_1 - 150) = 1500$ $-20V_1 + 258.5V_1 - 7050 = 1500$ $-20V_1 + 258.5V_1 - 7050 = 1500$ $238.5V_1 = 8550$ $V_1 = \frac{8550}{238.5} \approx 35.85 \text{ V}$

Substitute $V_1$ back to find $V_2$: $V_2 = 5.5(35.85) - 150 \approx 47.17 \text{ V}$

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Nodal voltage Matrix representation

In general, for any circuit, the node voltage equation can be written as $V/R=I$ $G \times V = I_S$

where

$G = \begin{bmatrix} G_{11} & -G_{12} & \dots & -G_{1n} \\ -G_{21} & G_{22} & \dots & -G_{nn} \\ \dots & \dots & \dots & \dots \\ G_{n1} & -G_{n2} & \dots & G_{nn} \end{bmatrix}$