Procedure (steps) for applying Nodal Analysis: –

1. Identify the total number of nodes.
2. One node selected as reference node and it is assigned to have ground (zero) potential and the remaining nodes called as nonreference node and we assign voltage designations to nonreference nodes. And at last check for supernode.
3. Develop the KCL equations for each nonreference node.
4. Solve the equations to find the unknown node voltages.

Note:- Apply both KCL and KVL to determine the node voltages.

Example

Example 1.  For the given network, find nodal voltages V₁ and V₂.

Solution:

As shown in the above Figure, given in the question, 1 V voltage source is connected between nodes 1 and 2, so node 1 and node 2 forms supernode. Thus this problem is based on supernode.

Step 1: – The total number of nodes is 3.

Step 2: – Node 0 is selected as reference node and it is assigned to have ground (zero) potential. The remaining node 1 and node 2 are considered as non-reference node shown in Figure 1. Here, node 1 and node 2 forms supernode.

Step 3 and Step 4: – Apply both KCL and KVL to determine the node voltages.

Apply KCL to supernode as shown in Figure 2,

\[2 + \frac{{({V_1} – 0)}}{1} + \frac{{({V_2} – 0)}}{1} + \frac{{({V_2} – 2)}}{1} = 0\]

${V_1} + 2{V_2} = 0$   ……(1)

Apply KVL to the loop having current I as shown in Figure 2,

\[ – {V_1} – 1 + {V_2} = 0\]

${V_1} – {V_2} = 1{\text{ }}$  ……(2)

Put Eq.(2) in Eq.(1), we get

\[3{V_2} + 1 = 0\]

\[{V_2} = – \frac{1}{3}{\text{ V}}\]

From Eq.(1), \[{V_1} + 2{V_2} = 0\]

\[{V_1} + 2\left( { – \frac{1}{3}} \right) = 0\]

\[{V_1} = \frac{2}{3}{\text{ V}}{\text{.}}\]

The post Supernode in electric circuits appeared first on ElectricalWorkbook.

Thevenin’s theorem states that for two terminal network N_A containing linear & bidirectional elements and independent sources is equivalent to a simple network containing an independent voltage source V_Th (called as Thevenin voltage) in series with the resistance R_Th (called as Thevenin resistance). The Thevenin equivalent circuit for network N_A shown below in Figure 1.

Procedure (steps) for applying Thevenin’s theorem: –

1. Calculation of Thevenin voltage V_Th
2. Calculation of Thevenin resistance R_Th for dc circuit and Thevenin impedance Z_Th for ac circuit. While calculating Thevenin resistance or impedance, all independent sources are set to zero (voltage source replaced by short circuits and current source replaced by open circuits).
3. Develop a Thevenin equivalent circuit and find the unknown parameter.

Thevenin theorem Example based on the DC circuit

Q For the given network, find the current in the following case R_L = 6 $\Omega $ and R_L = 36 $\Omega $ using Thevenin equivalent circuit.

Step 1: – Thevenin voltage V_Th calculation.

Remove R_L branch from the circuit and replace it with open circuit as shown in Figure 1 and let us assign the voltage V_Th (also called as Thevenin voltage) acorss the open circuited branch R_L. Apply KCL to node 1 shown in Figure 1,we get

\[\frac{{{V_1} – 32}}{4} – 2 + \frac{{{V_1}}}{{12}} = 0\]

\[{\text{or, }}{V_1} = 30{\text{ V}}\]

\[{\text{also }}{V_{{\text{Th}}}} = {V_1}\]

\[{\text{Thus, }}{V_{{\text{Th}}}} = {V_1} = 30{\text{ V}}\]

Step 2: – Thevenin Resistance R_Th calculation for ciruit shown in Figure 2, Replace 32 V voltage source as short circuit,

\[{R_{{\text{Th}}}} = (4||12) + 1\]

\[{\text{or, }}{R_{{\text{Th}}}} = \frac{{4 \times 12}}{{4 + 12}} + 1 = 3 + 1 = 4\]

\[{\text{Thus, }}{R_{{\text{Th}}}} = 4\Omega \]

Step 3: – The current through R_L using Thevenen equivalent circuit as shown in Figure 3 is

\[I = \frac{{{V_{{\text{Th}}}}}}{{{R_{{\text{Th}}}} + {R_L}}} = \frac{{30}}{{4 + {R_L}}}\]

when R_L = 6 $\Omega $,

\[I = \frac{{30}}{{4 + 6}} = 3{\text{ A}}\]

As shown in Figure 4, when R_L = 36 $\Omega $,

\[I = \frac{{30}}{{4 + 36}} = 0.75{\text{ A}}\] 

Thevenin theorem Example based on AC circuit

Q For the given network, find the Thevenin equivalent circuit between the terminals a and b.

Step 1:- Thevenin voltage ${{\mathbf{V}}_{{\text{Th}}}}$ calculation.

By applying voltage division to find ${{\mathbf{V}}_{{\text{Th}}}}$ as shown in Figure 1, we get

\[{{\mathbf{V}}_{{\text{Th}}}} = \frac{{20\angle 0^\circ \times 40}}{{j80 – j60 + 40}} = 17.92\angle – 26.5^\circ {\text{ V}}\]

Step 2: – Thevenin Impedance ${{\mathbf{Z}}_{{\text{Th}}}}$ calculation. Replace $20\angle 0^\circ V$ voltage source as short circuit as shown in Figure 2 gives

\[{{\mathbf{Z}}_{{\text{Th}}}} = (j80 – j60)||40)\]

\[= \frac{{(j80 – j60) \times 40}}{{(j80 – j60) + 40}}\]

\[= 17.88\angle 63.43^\circ {\text{ }}\Omega \]

Thus, ${{\mathbf{V}}_{{\text{Th}}}}$ = $17.92\angle – 26.5^\circ $ V, ${{\mathbf{Z}}_{{\text{Th}}}}$ = $17.88\angle 63.43^\circ {\text{ }}\Omega$

Step 3: – The Thevenin equivalent circuit is shown in Figure 3.

The post Thevenin’s Theorem in electric circuits appeared first on ElectricalWorkbook.

A mesh is a loop which does not contain any inner loop. Mesh analysis is only applicable to a planar network. A planar network is the one that can be drawn in a plane with no branches crossing one another.

Procedure (steps) for applying mesh analysis:

1. Identify the total number of meshes.
2. Assign the mesh currents.
3. Develop the KVL equation for each mesh.
4. Solve the equations to find the mesh currents.

Note:

• The total number of equations (e) required to solve the network with the help of mesh analysis is

e = b – (N – 1).

where, b is the total number of branches and N is the total number of nodes.

• The direction of mesh currents can be taken in any direction either clockwise or counter-clockwise. But clockwise direction results in a simpler analysis.

Examples

Example1. For the given network, find currents I₁ and I₂ using Mesh analysis.

Solution:

Let’s follow the Procedure for applying Mesh Analysis

Step 1: – The total number of meshes is 2.

Step 2: – The mesh currents I₁ and I₂ for mesh 1 and mesh 2 respectively as shown in Figure 1.

Step 3 and Step 4: – Apply KVL to mesh 1 as shown in Figure 2,

$ – 10 + 5{I_1} + 5({I_1} – {I_2}) = 0$

$2{I_1} – {I_2} = 2$                 ….(1)

Apply KVL to mesh 2 as shown in Figure 2,

$10{I_2} + 5({I_2} – {I_1}) = 0$

$15{I_2} – 5{I_1} = 0$

${I_1} = 3{I_2}$     ….(2)

Put Eq.(2) in Eq.(1), we get

$5{I_2} = 2$

\[{I_2} = \frac{2}{5}\]

from Eq.(2)

\[{I_1} =  3\left( {\frac{2}{5}} \right){\text{ = }}\frac{{ 6}}{5}{\text{ A}}\]

Thus,

\[{I_2} = \frac{2}{5}{\text{ A}};{I_1} = \frac{{ 6}}{5}{\text{ A}}{\text{.}}\]

Example 2. For the given network, write Mesh equations.

Solution:

Let’s follow the Procedure for applying Mesh Analysis

Step 1: – The total number of meshes is 3.

Step 2: – The mesh currents I₁, I₂ and I₃ for meshes 1, 2 and 3 are shown in Figure given in example 2.

Step 3: – Apply KVL to mesh 1,

 $- {E_1} + {R_1}{I_1} + {R_2}({I_1} – {I_3}) + {R_3}({I_1} – {I_2}) = 0{\text{}}$

Apply KVL to mesh 2,

\[ – {E_2} + {R_3}({I_2} – {I_1}) + {R_4}({I_2} – {I_3}) + {R_4}{I_2} = 0{\text{ }}\]

Apply KVL to mesh 3,

\[{R_6}{I_3} + {R_4}({I_3} – {I_2}) + {R_2}({I_3} – {I_1}) = 0{\text{}}\]

The post Mesh Analysis in Network theory appeared first on ElectricalWorkbook.

General theory of z parameters

In two port network, the terminal input and output voltages V₁ and V₂ can be expressed in terms of the terminal input and output currents I₁ and I₂.

As shown in Figure 1, the terminal voltages can be written in terms of the terminal currents as

${{\mathbf{V}}_1} = {{\mathbf{z}}_{{\text{11}}}}{{\mathbf{I}}_1} + {{\mathbf{z}}_{{\text{12}}}}{{\mathbf{I}}_2}$   …..(1)

${{\mathbf{V}}_2} = {{\mathbf{z}}_{{\text{21}}}}{{\mathbf{I}}_1} + {{\mathbf{z}}_{{\text{22}}}}{{\mathbf{I}}_2}$   …..(2)

From Eq.(1) and Eq.(2), it can be concluded that the terminal voltages are dependent variables and the terminal currents are independent variables. Also, using Eq.(1) and Eq.(2), the equivalent circuit represented as shown in Figure 2.

By using Eq.(1) and Eq.(2), can be written in matrix form as

$\left[ {\begin{array}{*{20}{c}}
{{{\mathbf{V}}_1}} \\
{{{\mathbf{V}}_2}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{{\mathbf{z}}_{{\text{11}}}}}&{{{\mathbf{z}}_{{\text{12}}}}} \\
{{{\mathbf{z}}_{{\text{21}}}}}&{{{\mathbf{z}}_{{\text{22}}}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{{\mathbf{I}}_1}} \\
{{{\mathbf{I}}_2}}
\end{array}} \right]$

The above matrix can be generalized as

$\left[ {\mathbf{V}} \right] = \left[ {\mathbf{z}} \right]\left[ {\mathbf{I}} \right]$

where [z] matrix,

$\left[ {\mathbf{z}} \right] = \left[ {\begin{array}{*{20}{c}}
{{{\mathbf{z}}_{{\text{11}}}}}&{{{\mathbf{z}}_{{\text{12}}}}} \\
{{{\mathbf{z}}_{{\text{21}}}}}&{{{\mathbf{z}}_{{\text{22}}}}}
\end{array}} \right]$

The elements of matrix [z] are called z parameters and each parameters having unit in ohm.

How to determine z parameters?

• The z parameters, z₁₁ and z₂₁ are obtained by open circuiting the port 2 i.e. I₂ = 0 along with connecting voltage source V₁ to port 1 as shown in Figure 3. Thus we get

\[{{\mathbf{z}}_{{\text{11}}}} = {\left. {\frac{{{{\mathbf{V}}_1}}}{{{{\mathbf{I}}_1}}}} \right|_{{{\mathbf{I}}_2} = 0}}\]

${{\mathbf{z}}_{{\text{11}}}}{\text{ is Open – circuit driving point input impedance}}{\text{.}}$

\[{{\mathbf{z}}_{{\text{21}}}} = {\left. {\frac{{{{\mathbf{V}}_2}}}{{{{\mathbf{I}}_1}}}} \right|_{{{\mathbf{I}}_2} = 0}}\]

${{\mathbf{z}}_{{\text{21}}}}{\text{ is Open – circuit forward transfer impedance }}{\text{.}}$

• The z parameters, z₂₂ and z₁₂ are obtained by open circuiting the port 1 i.e. I₁ = 0 along with connecting voltage source V₂ to port 2 as shown in Figure 4. Thus we get

\[{{\mathbf{z}}_{{\text{12}}}} = {\left. {\frac{{{{\mathbf{V}}_1}}}{{{{\mathbf{I}}_2}}}} \right|_{{{\mathbf{I}}_1} = 0}}\]

\[{{\mathbf{z}}_{{\text{12}}}}{\text{ is Open – circuit reverse transfer impedance}}{\text{.}}\]

\[{{\mathbf{z}}_{22}} = {\left. {\frac{{{{\mathbf{V}}_2}}}{{{{\mathbf{I}}_2}}}} \right|_{{{\mathbf{I}}_1} = 0}}\]

${{\mathbf{z}}_{22}}{\text{ is Open – circuit driving point output impedance}}{\text{.}}$

Note:- For z parameters, the terminal voltage V₁ and V₂ are function of terminal currents I₁ and I₂ as

\[\left( {{{\mathbf{V}}_1},{{\mathbf{V}}_2}} \right) = {\text{f }}\left( {{{\mathbf{I}}_1},{{\mathbf{I}}_2}} \right)\]

Condition of symmetry

A two port network is said to be symmetrical if the ratio of response to excitation remains same at both ports independently with respect to the defined circuit conditions such as open circuit or short circuit. Thus, we can say

\[{\left. {\frac{{{{\mathbf{V}}_1}}}{{{{\mathbf{I}}_1}}}} \right|_{{{\mathbf{I}}_2} = 0}} = {\left. {\frac{{{{\mathbf{V}}_2}}}{{{{\mathbf{I}}_2}}}} \right|_{{{\mathbf{I}}_1} = 0}}\]

\[{\text{or, }}{{\mathbf{z}}_{11}} = {{\mathbf{z}}_{22}}\]

Condition of reciprocity

A two port network is said to be reciprocal if the ratio of response to excitation remains same even when the position of response to excitation are interchanged. Thus, we can say

\[{\left. {\frac{{{{\mathbf{V}}_1}}}{{{{\mathbf{I}}_2}}}} \right|_{{{\mathbf{I}}_1} = 0}} = {\left. {\frac{{{{\mathbf{V}}_2}}}{{{{\mathbf{I}}_1}}}} \right|_{{{\mathbf{I}}_2} = 0}}\]

\[{\text{or, }}{{\mathbf{z}}_{12}} = {{\mathbf{z}}_{21}}\]

Applications

• Use for filters Combination.
• design of impedance-matching networks
• design of power distribution networks.

The post z parameters / Impedance Parameters / open circuit parameters in two port network appeared first on ElectricalWorkbook.

Procedure (steps) for applying mesh analysis:

1. Identify the total number of meshes.
2. Assign the mesh currents and check for supermesh in the circuit.
3. If supermesh found, develop the KVL equation for it.
4. Solve the equations to find the mesh currents.

Example

Example 1. For the given network, find current I using Mesh analysis.

As shown above, Figure is given in example 1, 2 A current source is connected between meshes 1 and 2 so this problem is based on supermesh.

Step 1: – The total number of meshes is 2.

Step 2: – Let us assign mesh currents I₁ and I₂ for meshes 1 and 2 respectively as shown in Figure 1. As shown in Figure 1, 2 A current source should be removed from the circuit because 2 A current source is connected between meshes 1 and 2.

Figure 1.

Step 3: – The reduced circuit having supermesh shown in Figure 2.

Figure 2.

Apply KVL to supermesh

\[{\text{ – 4 + }}{I_1} + {I_2} + 2 = 0\]

${I_1} + {I_2} = 2$             ….(1)

Apply KCL to node 0,

\[{I_2} – {I_1} = 2{\text{ }}\]

${I_2} = {I_1} + 2$             ….(2)

Put equation (2) in equation (1), we get

\[{I_1} + {I_1} + 2 = 2\]

\[2{I_1} + 2 = 2\]

${I_1} = 0{\text{ A }}$         ….(3)

From Equation (1) ,

\[{I_1} + {I_2} = 2\]

Put equation (3) in equation (1), we get

\[0 + {I_2} = 2\]

\[{I_2} = 2\]

also \[I = {I_2}\]

Therefore,

\[I = 2{\text{ A}}{\text{.}}\]

The post Super mesh Analysis in Network theory appeared first on ElectricalWorkbook.

Voltage and Current Relationship

Let us consider a circuit having resitance R supplied by a ac source voltage $v(t)$ as shown in Figure 1 is

$v(t) = {V_m}\sin \omega t$   ….(1)

The current passes throgh resistance R will be

\[i(t) = \frac{{v(t)}}{R} = \frac{{{V_m}}}{R}\sin \omega t = {I_m}\sin \omega t\]

Thus,

$i(t) = {I_m}\sin \omega t$   ….(2)

Phasor Diagram

The phasor representation of Equations (1) and (2), is shown in Figure 2, can be written and relate in phasor terms as

\[{\mathbf{I}} = \frac{{\mathbf{V}}}{R}\]

or,

\[{\mathbf{V}} = {\mathbf{I}}R\]

Equations (1) and (2) shows that both current and voltage are in phase so waveforms will look as shown in Figure 3.

Power calculation

The instantaneous power $p(t)$ is defined by the product of instantaneous voltage $v(t)$ and instantaneous current $i(t)$.

The instantaneous power waveform as shown in Figure 3. And according to the definition, instantaneous power $p(t)$ write as

$p(t) = v(t)i(t)$   ….(3)

Using Equation (3), the average power is defined as

\[{P_{av}} = \frac{1}{T}\int\limits_0^T {p(t)dt} \]

Also,

\[{P_{av}} = \frac{1}{T}\int\limits_0^T {v(t)i(t)dt} \]

\[ = \frac{1}{T}\int\limits_0^T {\left( {{V_m}\sin \omega t} \right)\left( {{I_m}\sin \omega t} \right)dt} \]

\[ = \frac{{{V_m}{I_m}}}{T}\int\limits_0^T {{{\sin }^2}\omega tdt} \]

\[ = \frac{{{V_m}{I_m}}}{{2T}}\int\limits_0^T {\left( {1 – \cos 2\omega t} \right)dt} \]

\[ = \frac{{{V_m}{I_m}}}{2}\]

\[ = \frac{{{V_m}}}{{\sqrt 2 }}\frac{{{I_m}}}{{\sqrt 2 }}\]

\[ = {V_{rms}}{I_{rms}}\]

Thus,

\[{P_{av}} = {V_{rms}}{I_{rms}}.\]

Also, Figure 3 shows that the frequency of instantaneous power ${f_p}$ is twice the frequency of instantaneous voltage ${f_v}$ or the frequency of instantaneous current ${f_i}$ as

\[{f_p} = 2{f_i}{\text{ }}or{\text{ }}2{f_v}\]

The post AC Supply to Pure Resistor appeared first on ElectricalWorkbook.

General theory of y parameters

In two port network, the terminal input and output voltages V₁ and V₂ can be expressed in terms of the terminal input and output currents I₁ and I₂.

As shown in Figure 1, the terminal voltages can be written in terms of the terminal currents as

${{\mathbf{I}}_1} = {{\mathbf{y}}_{{\text{11}}}}{{\mathbf{V}}_1} + {{\mathbf{y}}_{{\text{12}}}}{{\mathbf{V}}_2}$  ….(1)

${{\mathbf{I}}_2} = {{\mathbf{y}}_{{\text{21}}}}{{\mathbf{V}}_1} + {{\mathbf{y}}_{{\text{22}}}}{{\mathbf{V}}_2}$   ….(2)

From Eq.(1) and Eq.(2), it can be concluded that the terminal currents are dependent variables and the terminal voltages are independent variables. Also, using the Eq.(1) and Eq.(2), the equivalent circuit represented as shown in Figure 2.

By using Eq.(1) and Eq.(2), can be written in matrix form as

$\left[ {\begin{array}{*{20}{c}}
{{{\mathbf{I}}_1}} \\
{{{\mathbf{I}}_2}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{{\mathbf{y}}_{{\text{11}}}}}&{{{\mathbf{y}}_{{\text{12}}}}} \\
{{{\mathbf{y}}_{{\text{21}}}}}&{{{\mathbf{y}}_{{\text{22}}}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{{\mathbf{V}}_1}} \\
{{{\mathbf{V}}_2}}
\end{array}} \right]$

The above matrix can be generalized as

$\left[ {\mathbf{I}} \right] = \left[ {\mathbf{y}} \right]\left[ {\mathbf{V}} \right]$

where [y] matrix,

$\left[ {\mathbf{y}} \right] = \left[ {\begin{array}{*{20}{c}}
{{{\mathbf{y}}_{{\text{11}}}}}&{{{\mathbf{y}}_{{\text{12}}}}} \\
{{{\mathbf{y}}_{{\text{21}}}}}&{{{\mathbf{y}}_{{\text{22}}}}}
\end{array}} \right]$

The elements of matrix [y] are called y parameters and each parameters having unit in mho.

How to determine y parameters?

• The y parameters y₁₁ and y₂₁ are obtained by short circuiting the port 2 i.e. V₂ = 0 along with connecting current source I₁ to port 1 as shown in Figure 3. Thus we get

\[{{\mathbf{y}}_{{\text{11}}}} = {\left. {\frac{{{{\mathbf{I}}_1}}}{{{{\mathbf{V}}_1}}}} \right|_{{{\mathbf{V}}_2} = 0}}\]

${{\mathbf{y}}_{{\text{11}}}}{\text{is Short – circuit driving point input admittance}}{\text{.}}$

\[{{\mathbf{y}}_{{\text{21}}}} = {\left. {\frac{{{{\mathbf{I}}_2}}}{{{{\mathbf{V}}_1}}}} \right|_{{{\mathbf{V}}_2} = 0}}\]

${{\mathbf{y}}_{{\text{21}}}}{\text{ is Short – circuit forward transfer admittance}}{\text{.}}$

• The y parameters y₂₂ and y₁₂ are obtained by short circuiting the port 1 i.e. V₁ = 0 along with connecting current source I₂ to port 2 as shown in Figure 4. Thus we get

\[{{\mathbf{y}}_{{\text{12}}}} = {\left. {\frac{{{{\mathbf{I}}_1}}}{{{{\mathbf{V}}_2}}}} \right|_{{{\mathbf{V}}_1} = 0}}\]

${{\mathbf{y}}_{{\text{12}}}}{\text{ is Short – circuit reverse transfer admittance}}{\text{.}}$

\[{{\mathbf{y}}_{22}} = {\left. {\frac{{{{\mathbf{I}}_2}}}{{{{\mathbf{V}}_2}}}} \right|_{{{\mathbf{V}}_1} = 0}}\]

${{\mathbf{y}}_{22}}{\text{ is Short – circuit driving point output admittance}}{\text{.}}$

Note:-  For y parameters, the terminal currents I₁ and I₂ are function of terminal voltages V₁ and V₂ as

$\left( {{{\mathbf{I}}_1},{{\mathbf{I}}_2}} \right) = {\text{f }}\left( {{{\mathbf{V}}_1},{{\mathbf{V}}_2}} \right)$

Condition of symmetry

A two port network is said to be symmetrical if the ratio of response to excitation remains same at both ports independently with respect to the defined circuit conditions such as open circuit or short circuit. Thus, we can say

\[{\left. {\frac{{{{\mathbf{I}}_1}}}{{{{\mathbf{V}}_1}}}} \right|_{{{\mathbf{V}}_1} = 0}} = {\left. {\frac{{{{\mathbf{I}}_2}}}{{{{\mathbf{V}}_2}}}} \right|_{{{\mathbf{V}}_2} = 0}}\]

\[{\text{or, }}{{\mathbf{y}}_{11}} = {{\mathbf{y}}_{22}}\]

Condition of reciprocity

A two port network is said to be reciprocal if the ratio of response to excitation remains same even when the position of response to excitation are interchanged. Thus, we can say

\[{\left. {\frac{{{{\mathbf{I}}_1}}}{{{{\mathbf{V}}_2}}}} \right|_{{{\mathbf{V}}_1} = 0}} = {\left. {\frac{{{{\mathbf{I}}_2}}}{{{{\mathbf{V}}_1}}}} \right|_{{{\mathbf{V}}_2} = 0}}\]

\[{\text{or, }}{{\mathbf{y}}_{12}} = {{\mathbf{y}}_{21}}\]

Applications

• Use for filters Combination.
• design of impedance-matching networks
• design of power distribution networks.

The post Admittance Parameters/ y parameters / short circuit parameters in two port network appeared first on ElectricalWorkbook.

Norton’s theorem states that for two terminal network N_A containing linear & bidirectional elements and independent sources is equivalent to a simple network containing an independent current source I_N (called as Norton current) in parallel with the resistance R_N (called as Norton resistance). The Norton equivalent circuit for network N_A shown below in Figure 1.

Procedure (steps) for applying Norton’s theorem: –

1. Calculation of Norton current I_N
2. Calculation of Norton resistance R_N for dc circuit and Norton impedance Z_N for ac circuit. The Thevenin resistance equal to Norton resistance and the Thevenin impedance equal to Norton impedance. While calculating Norton resistance or impedance, all independent sources are set to zero (voltage source replaced by short circuits and current source replaced by open circuits).
3. Develop the Norton equivalent circuit and find the unknown parameter.

Norton theorem Example based on the DC circuit

Q For the given network, find the current I passes through 5 $\Omega $ using Norton equivalent circuit.

Step 1: – Norton current I_N calculation

Remove 5 $\Omega $ branch from the circuit and replace it with a short circuit as shown in Figure 1 and let us assign the current I_N (also called as Norton current) passes through the short circuit. By applying the current divider rule in Figure 1 gives

\[{I_{\text{N}}} = \frac{{20}}{{5 + 2 + 3}}\left( 5 \right) = 10{\text{ A}}{\text{.}}\]

Step 2: – Norton Resistance R_N calculation

Replace 20 A current source as open circuit as shown in Figure 2, Thus R_N  for the circuit is

\[{R_{\text{N}}} = 3 + 2 + 5 = 10{\text{ }}\Omega \]

Step 3: – The Norton equivalent circuit as shown in Figure 3, and the current through 5 $\Omega $ branch calculated using current divider rule gives

\[I = \frac{{10}}{{5 + 10}}\left( 10 \right) = 6.67{\text{ A}}{\text{.}}\]

Norton theorem Example based on AC circuit

Q For the given network, find the Norton equivalent circuit between the terminals a and b.

Step 1: – Norton current ${{\mathbf{I}}_{{\text{N}}}}$ calculation

By applying Ohm’s law to find ${{\mathbf{I}}_{{\text{N}}}}$ as shown in Figure 1, we get,

\[{{\mathbf{I}}_{{\text{N}}}} = \frac{{10\angle 0^\circ }}{{10 – j10}} = \frac{1}{{1 – j1}}{\text{ = }}\frac{1}{{\sqrt 2 }}\angle 45^\circ {\text{ A}}\]

Step 2: – Norton Impedance ${{\mathbf{Z}}_{{\text{N}}}}$ calculation

Replace ${\text{1}}0\angle 0^\circ$ V voltage source as short circuit as shown in Figure 2 gives

\[{{\mathbf{Z}}_{\text{N}}} = ((10 – j10)||j20) = \frac{{(10 – j10) \times j20}}{{(10 – j10) + j20}} = 20{\text{ }}\Omega \]

\[{\text{Thus, }}{{\mathbf{I}}_{{\text{N}}}}{\text{ = }}\frac{1}{{\sqrt 2 }}\angle 45^\circ {\text{ A}},{\text{ }}{{\mathbf{Z}}_{\text{N}}} = 20{\text{ }}\Omega \]

Step 3: – The Norton equivalent circuit as shown in Figure 3

The post Norton’s Theorem in electric circuits appeared first on ElectricalWorkbook.

Superposition theorem states that in a linear bilateral network containing more than one independent source, the response in any element is the sum of the response obtained with one source acting at a time and other source being deactivated. Deactivation means all the independent sources are replaced by their internal resistances i.e. voltage source replaced by a short circuit and current source replaced by an open circuit while retaining all the dependent sources as they are.

Procedure (steps) for applying Superposition Theorem:-

1. Select any one independent source and do the calculation for voltage or current due to this source.
2. Repeat step-1 for each independent source.
3. Algebraically add the results of each source.

Superposition theorem Example based on the DC circuit

Q For the given network, find the current I using superposition theorem.     

Solution:

Step 1: – Consider 12 V voltage source, replace 24 V voltage source as a short circuit and 3 A current source as an open circuit in Figure 1.

Thus,

\[{I_1} = \frac{{12}}{6} = 2{\text{ A}}\]

Step 2: – Consider 24 V voltage source, replace 12 V voltage source as a short circuit and 3 A current source as an open circuit, shown in Figure 2.

Then, Apply KCL to node 1, we get

\[\frac{{{V_1} – 24}}{{12}} + \frac{{{V_1}}}{4} + \frac{{{V_1}}}{3} = 0\]

\[{\text{or, 8}}{V_1} = 24\]

\[{\text{or, }}{V_1} = 3{\text{ V}}\]

\[{I_2} = \frac{{0 – {V_1}}}{3} = \frac{{ – 3}}{3} = – 1\]

\[{\text{Thus, }}{I_2} = – 1{\text{ A}}\]

Repeat step 2: – Consider 3 A current source, replace 24 V voltage source as short circuit and 12 V voltage source as short circuit, shown in Figure 3.

Then, Apply KCL to node 1,we get

\[\frac{{{V_1}}}{4} + \frac{{{V_1}}}{3} + \frac{{{V_1} – {V_2}}}{{12}} = 0\]

${\text{or, }}3{V_2} = {\text{10}}{V_1}$  …..(1)

Apply KCL to node 2,we get

\[\frac{{{V_2}}}{8} + \frac{{{V_2} – {V_1}}}{4} – 3 = 0\]

\[{\text{or, }}\frac{{{V_2}}}{8} + \frac{{{V_2} – {V_1}}}{4} = 3\]

${\text{or, }}3{V_2} – 2{V_1} = 24$  …..(2)

Put Eq.(1) in Eq.(2), we get

\[10{V_1} – 2{V_1} = 24{\text{ }}\]

\[{\text{or, }}{V_1} = 3{\text{ V}}\]

\[{I_3} = \frac{{{V_1}}}{3} = \frac{3}{3} = 1\]

${\text{Thus, }}{I_3} = 1{\text{ A}}$

Step 4:- By applying superposition theorem,

\[I = {I_1} + {I_2} + {I_3}\]

\[{\text{or, }}I = 2 – 1 + 1\]

\[I = 2{\text{ A}}\]

Superposition theorem Example based on AC circuit

 Q For the given network, find the voltage V_x using superposition theorem.

Step 1: – Consider $1\angle 0^\circ$ A current source, replace $1\angle 0^\circ$ V voltage source as short circuit, shown in Figure 1.

Then, by applying current division to find ${{\mathbf{I}}_1}$, we get

\[{{\mathbf{I}}_1} = \frac{{1\angle 0^\circ \times j}}{{1 + j}}\]

\[{{\mathbf{V}}_{x1}} = {{\mathbf{I}}_1} \times 1\]

\[{\text{or, }}{{\mathbf{V}}_{x1}} = \frac{{1\angle 0^\circ \times j}}{{1 + j}} \times 1\]

\[{\text{or, }}{{\mathbf{V}}_{x1}} = \frac{j}{{1 + j}}{\text{ V}}\]

Step 2: – Consider $1\angle 0^\circ$ V voltage source, replace $1\angle 0^\circ$ A current source as open circuit, shown in Figure 2.

Then apply voltage division rule we get,

\[{{\mathbf{V}}_{x2}} = \frac{{1\angle 0^\circ \times 1}}{{1 + j1}}\]

\[{\text{or, }}{{\mathbf{V}}_{x2}} = \frac{1}{{1 + j}}{\text{ V}}\]

Step 3: – By applying superposition theorem,

\[{{\mathbf{V}}_x} = {{\mathbf{V}}_{x1}} + {{\mathbf{V}}_{x2}}\]

\[{\text{or, }}{{\mathbf{V}}_x} = \frac{j}{{1 + j}} + \frac{1}{{1 + j}} = \frac{{1 + j}}{{1 + j}} = 1\angle 0^\circ \]

\[{\text{Thus, }}{{\mathbf{V}}_x} = 1\angle 0^\circ {\text{ V}}\]

Limitations

1. Superposition theorem is not applicable in case of Nonlinear networks and for the parameters which show nonlinear relationship like power.
2. When it applied on a circuit, direction (sign) of currents supplied by each source should be noticed.

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Voltage and Current Relationship

Let us consider a circuit having inductor L supplied by an ac source voltage $v(t)$ which produce a sinusoidal current $i(t)$ as shown in Figure 1 is

$i(t) = {I_m}\sin \omega t$  ….(1)

The general expression that relates inductor voltage and current is

\[v(t) = L\frac{{di(t)}}{{dt}}{\text{ }}…(2)\]

Put equation (1) in equation (2),

\[v(t) = L\frac{d}{{dt}}\left( {{I_m}\sin \omega t} \right)\]

\[ = \omega L{I_m}\cos \omega t\]

\[ = \omega L{I_m}\sin \left( {\omega t + 90^\circ } \right)\]

\[ = j\omega L{I_m}\sin \omega t\]

Thus,

$v(t) = j\omega L{I_m}\sin \omega t$   ….(3)

we know,

\[v(t) = \omega L{I_m}\cos \omega t = {V_m}\cos \omega t\]

where, ${V_m} = \omega L{I_m}$

Phasor diagram

The phasor representation of Equations (1) and (3), is shown in Figure 2, can be written and relate in phasor terms as

\[{\mathbf{V}} = j\omega L{\mathbf{I}} = j{X_L}{\mathbf{I}}\]

where ${X_L} = \omega L$ and called as inductive reactance. Equations (1) and (2) shows that current lags voltage by phase angle of ${\text{90}}^\circ$ so waveforms will look as shown in Figure 3.

Power calculation

The instantaneous power $p(t)$ is defined by the product of instantaneous voltage $v(t)$ and instantaneous current $i(t)$.

The instantaneous power $p(t)$ waveform shown in Figure 3. And according to the definition, instantaneous power $p(t)$ write as

$p(t) = v(t)i(t)$   ….(4)

Using Equation (4), the average power is defined as

\[{P_{av}} = \frac{1}{T}\int\limits_0^T {p(t)dt} \]

Also,

\[{P_{av}} = \frac{1}{T}\int\limits_0^T {v(t)i(t)dt} \]

\[ = \frac{1}{T}\int\limits_0^T {\left( {{V_m}\cos \omega t} \right)\left( {{I_m}\sin \omega t} \right)dt} \]

\[ = \frac{{{V_m}{I_m}}}{T}\int\limits_0^T {\sin \omega t\cos \omega tdt} \]

\[ = \frac{{{V_m}{I_m}}}{{2T}}\int\limits_0^T {2\sin \omega t\cos \omega tdt} \]

\[ = \frac{{{V_m}{I_m}}}{{2T}}\int\limits_0^T {\sin 2\omega tdt} \]

\[ = 0\]

Thus,

\[{P_{av}} = 0.\]

Also, Figure 3 shows that the frequency of instantaneous power ${f_p}$ is twice the frequency of instantaneous voltage ${f_v}$ or the frequency of instantaneous current ${f_i}$ as

\[{f_p} = 2{f_i}{\text{ }}or{\text{ }}2{f_v}\]

Note:- In the first half cycle of the instantaneous power $p(t)$, the inductor takes the energy from the source and in the second half cycle of the instantaneous power $p(t)$, inductor delivers the energy back to the source.

The post AC supply to Pure Inductor appeared first on ElectricalWorkbook.