UNIT-2

DC Circuits

2.1 DC Circuit analysis by mesh current method and Nodal voltage method

LOOP ANALYSIS

For Loop 1 with il

For Loop 2 with i2

For Loop 3 with i3

Example 1. For the circuits given below write the voltage equations:

Solution: Let current i1be in loop 1 current and i2 for loop 2 

Solution:

For loop 1 

For loop 2 

NODE ANALYSIS

For these we assume every node as a voltage point and write the current equation for every element. For current source, current entering is negative.

For node V1

For node V2

For V

Example: Using nodal analysis find voltage across 5resistor.

Solution:

For V1

      1

For V2

     2

Solving 1 and 2:

For 5 voltage =

= -50.9 + 57.27

= 6.37V

2.2 Superposition theorem, Thevenin’s theorem and maximum power transfer theorem –Application to simple DC circuits

SUPERPOSITION THEOREM

1. This is only applicable to circuits with linear elements.
2. If two or more than two independent sources (voltage or current) are operating in the circuit than voltage across any element or current through any element is sum of current and voltages due to individual sources.

Question 1. Find the current through resistance.

Solution:

1 = 0

2   =

1 + 2 

Special Case:

Since two voltage sources with different magnitude in parallel which cannot be connected as in single branch two different current is not possible (if 5V than I = zero).

Question:

1 =

2 =

=

1 + 2

=

THEVENIN’S AND NORTON’S THEOREM

Thevenin’s equivalent of A

Norton’s equivalent of A

sc = Vth/Rth

• Norton’s equivalent is obtained by source conversion of thevenin’s equivalent circuit.

CONDITIONS FOR APPLICATION

• For network A:

1. Network A should contain linear elements.
2. Network A can have independent and dependent current and voltage source.
3. If network A has dependent source than controlling parameter must lie in network A itself.
4. Network A should not have any source coupling and magnetic coupling.

• For network B:

1. It can have linear and non-linear elements.
2. It can have dependent and independent voltage and current sources.
3. It should not have any source and magnetic coupling with network A.

     Method for finding Rth :

Firstly, open circuit terminal A and B.

1. If network is operating with only independent sources:

1. Make all sources zero in network A.
2. Find out the equivalent resistance across terminal A and B.

2.     If network A is operating with independent and dependent sources:

1. Make all independent sources zero in network A.
2. Connect a generation between A and B.

3.     If network is operating with only dependent sources:

Connect generation between A and B

     Method for Vth:

First open circuit terminal A and B.

Find out the voltage between A and B this is Vth

     Method for Isc:

1. Isc =
2. Remove network B and S.C. The terminal A and B and current from terminal A to B Isc.

Question:

Answer:

Finding Isc from circuit directly:

By KCL,

Question:

Answer

Also, clear from circuit that Vth = 1V.

By applying KVL we get,

1-3Isc=0

Isc=A

Que:

Ans;

Rth=3k+2k=5k

By applying KVL we get

Therefore,

Question:

Solution: For Rth

By KCL,

But,

By KVL,

Question:

Solution: Since, no independent source is present so,

Isc = 0

And we know that,

Since Rth cannot be zero

But

Question: Find out the Norton’s equivalent

Solution:

Since, there is no significance of current source

MAXIMUM POWER TRANSFER THEOREM

Where Rth is Thevenin’s equivalent resistance across a and b.

Maximum power is absorbed by ZL  when

Condition:

Comparing real and imaginary parts

OR

Maximum power absorbed by ZL is

Question: Find out the value of load resistance if power absorbed is maximum.

Solution: find Thevenin’s equation

Question: Find maximum power delivered is RL if its value is

1. 16Ω
2. Ω
3. 60Ω
4. 20Ω

Solution

Therefore,