DC Circuits

 DC Circuits

Topics Covered :

1)Kirchhoff's Laws

a) Kirchhoff's Voltage Law (KVL)

b) Kirchhoff's Current Law (KCL)

2) Mesh Analysis & Super Mesh

3) Nodal Analysis & Super Node

4) Delta to Star Transformation

5) Star to Delta Transformation

6) Theorems

a) Superposition Theorem

b) Thevenin's Theorem

c) Norton's Theorem

d) Maximum Power Transfer Theorem

1.Kirchhoffs Laws

a) Kirchhoff's Voltage Law

It states that for any closed loop in a circuit, the sum of the potential differences across all components is zero

b) Kirchhoff's Current Law

The algebraic sum of branch currents flowing into and out of a node is equal to zero.

2. Mesh Analysis & Super Mesh

Mesh : Mesh analysis (or the mesh current method) is a method that is used to solve planar circuits for the currents (and indirectly the voltages) at any place in the electrical circuit. Planar circuits are circuits that can be drawn on a plane surface with no wires crossing each other.

Steps to solve a Mesh problem

step 1 : Give Direction to the Current flowing in Loops

step 2 : Apply KVL and get no.of equations

step 3 : Put Equations in Simultaneous Mode of scientific Calculator and Solve

step 4 : You will Now get value of \(I_3\) \(I_2\) \(I_3\) ......\(I_n\)

Example :

Apply Mesh to Following Circuit :

Apply KVL to loop 1

5\(I_1\)+2\(I_1\)-2\(I_2\)-25=0

7\(I_1\)-2\(I_2\)+0\(I_3\)=25 --- Eqn 1

Apply KVL to loop 2

2\(I_2\)-2\(I_1\)+3\(I_2\)+4\(I_2\)-4\(I_3\)=0

-2\(I_1\)+9\(I_2\)-4\(I_3\)=0 --- Eqn 2

Apply KVL to loop 3

4\(I_3\)-\(I_2\)+6\(I_3\)=0

0\(I_1\)-4\(I_2\)+10\(I_3\)=0 --- Eqn 3

Solve Eqn 1,2,3 Simultaneously we get,

7\(I_1\)-2\(I_2\)+0\(I_3\)=25 --- Eqn 1

2\(I_1\)+9\(I_2\)-4\(I_3\)=0 --- Eqn 2

0\(I_1\)-4\(I_2\)+10\(I_3\)=0 --- Eqn 3

Ans -

\(I_1\)= 3.315412 Amps

\(I_2\)= -0.89605 Amps

\(I_3\)= -0.35842 Amps

Super Mesh : A supermesh occurs when a current source is contained between

two essential meshes

Example : Supermesh -

Find current through 5Ω Resistor

Apply KVL to Loop 1

10\(I_1\)-10\(I_2\)+5\(I_1\)-5\(I_3\)=50

15\(I_1\)-10\(I_2\)-5\(I_3\)=50 --- Eqn 1

Mesh 2 & Mesh 3 Shares Same current Source due to which a Super mesh is formed

Writing Current Eqution for Super Mesh

\(I_2\)-\(I_3\)=2 --- Eqn 2

Applying KVL to Supermesh

10\(I_2\)-10\(I_1\)+2\(I_2\)+1\(I_3\)+5\(I_3\)-5\(I_1\)=0

-15\(I_1\)+12\(I_2\)+6\(I_3\)=0 --- Eqn 3

Solving Eqn 1,2,3 Simultaneously We get

\(I_1\) = 20 Amps

\(I_2\) = 17.33 Amps

\(I_3\) = 15.33 Amps

hence \(I_{5\Omega}\)=\(I_1\)-\(I_3\)

=20-15.33= 4.67 Amps

∴ \(I_{5\Omega}\)= 4.67 Amps

3. Delta To Star Transformation

Formulae

\( R_{1} = \frac{R_{b} R_{c}}{R_{a} + R_{b} + R_{c}} \)

\( R_{2} = \frac{R_{a} R_{c}}{R_{a} + R_{b} + R_{c}} \)

\( R_{3} = \frac{R_{a} R_{b}}{R_{a} + R_{b} + R_{c}} \)

4. Star to Delta Transformation

Formulae

\( R_{a} = R_{2} + R_{3} + \frac{R_{2} R_{3}}{R_{1}} \)

\( R_{b} = R_{1} + R_{3} + \frac{R_{1} R_{3}}{R_{2}} \)

\( R_{c} = R_{1} + R_{2} + \frac{R_{1} R_{2}}{R_{3}} \)

5.Theorems
a. Superposition Theorem

The voltage or current in any one branch of a linear, passive network can be

determined by considering the effects of each independent source separately.