Unit-1
DC Circuits
Resistor
Control the flow of current
V=IR
P=I2R
Behaviour of Resistance:-
t=0- t=0 t=0+ t=∞
R R R R
Inductor
- Active - FET, diode
- Passive - R, L, C
Storage of energy
Based on electromagnetic induction
Faraday’s Law
Φ∝ i
Φ = L.i
Relaxed inductor I0 = 0
Current Carrying Inductor:
L = L1+L2
Capacitor:
Charge storing device (voltage)
Q ∝ V
Q = C V
Relaxed capacitor
charged capacitor
Electrostatic Energy
Inductor do not respond any change as far as current is concerned.
Capacitor does not respond any change as far as voltage is concerned.
Definition of Open circuit and short circuit
R=∞
V=0 or V≠ 0
I=0
Short Circuit:
V = 0
R = 0
I may or may not be zero
Categories of sources
- Voltage source and current source
- A.C. Source and D.C. Source
- Dependent source and independent source
- Constant and function of time source
few either
AC or DC
dc
Equivalent circuit diagram of
- Voltage Source: A voltage source always have a resistance in series.
Ideal source
R = 0
Practical = 2 - 5Ω
I = 0
2. Current Source: Always have R parallel with the source.
Ideal and Practical Sources (Independent Sources only)
Ideal and practical Voltage and Current source:
A voltage source is a device which provides a constant voltage to load at any instance of time and is independent of the current drawn from it. This type of source is known as an ideal voltage source. Practically, the ideal voltage source cannot be made. It has zero internal resistance. It is denoted by this symbol.
Fig: Voltage source symbol
Ideal Voltage Source
Fig: Ideal Voltage Source
The graph represents the change in voltage of the voltage source with respect to time. It is constant at any instance of time.
Voltage sources that have some amount of internal resistance are known as a practical voltage source. Due to this internal resistance, voltage drop takes place. If the internal resistance is high, less voltage will be provided to load and if the internal resistance is less, the voltage source will be closer to an ideal voltage source. A practical voltage source is thus denoted by a resistance in series which represents the internal resistance of source.
Practical Voltage source
Fig: Practical Voltage source
The graph represents the voltage of the voltage source with respect to time. It is not constant but it keeps on decreasing as the time passes.
Current source
A current source is a device which provides the constant current to load at any time and is independent of the voltage supplied to the circuit. This type of current is known as an ideal current source; practically ideal current source is also not available. It has infinite resistance. It is denoted by this symbol.
Ideal Current source
Fig: Ideal Current source
The graph represents the change in current of the current source with respect to time. It is constant at any instance of time.
Why ideal Current source has infinite resistance?
A current source is used to power a load, so that load will turn on. We try to supply 100% of the power to load. For that, we connect some resistance to transfer 100% of power to load because the current always takes the path of least resistance. So, in order for current to go to the path of least resistance, we must connect resistance higher than load. This is why we have the ideal current source to have infinite internal resistance. This infinite resistance will not affect voltage sources in the circuit.
Practical Current source
Practically current sources do not have infinite resistance across there but they have a finite internal resistance. So the current delivered by the practical current source is not constant and it is also dependent somewhat on the voltage across it.
A practical current source is represented as an ideal current source connected with resistance in parallel.
Fig; Practical Current source
The graph represents the current of the current source with respect to time. It is not constant but it also keeps on decreasing as the time passes.
The algebraic sum of currents meeting at a junction or node in a electric circuit is zero or the summation of all incoming current is always equal to summation of all outgoing current in an electrical network.
Explanation
Assuming the incoming current to be positive and outgoing current negative we have
Ie incoming current = ∑ outgoing current thus, the above Law can also be stated as the sum of current flowing towards any junction in an electric circuit is equal to the sum of currents flowing away from that junction
Kirchhoff’s Voltage Law (KVL)
Statement : the algebraic summation of all Voltage in any closed circuit or mesh of loop zero.
Ie ∑ Voltage in closed loop = 0 the summation of the Voltage rise (voltage sources) is equal to summation of the voltage drops around a closed loop in 0 circuit for explanation from here
Determination of sigh and direction of currents (Don’t write in exams just for understanding)
Current entering a resistor is +ve and leaving should be –ve
Now
Potential Rise Potential Drop
We are reading from +V to –V we are reading from –V to +V
potential drops potential rise
-V +V
Given Circuit
First identify no of loops and assign direction of current flowing in loop
Note : no of loops in circuit = No, of unknown currents = no, of equations in the circuit
Note : keep loop direction and current direction same ie either clockwise or anticlockwise for all loops I1 I2
Now according to direction of direction assign signs (+ve to –ve) to the resistors
Note : voltage sources (V) polarities does not change is constant.
Note: for common resistor between 2 loops appearing in the circuit like R3 give signs according to separate loops as shown
When considering only loop no 1 (+ R3 - )
- B
Now consider diagram A and write equations
Two loops two unknown currents two equation
Apply KVL for loop ① [B. Diagram ]
(+ to drop -) = - sign and (- to rise +) = + sign
for drop = -sign
for rise = + sign
-
-() R2 is considered because in R3,2 currents are flowing and and we have taken () because we are considering loop no 1 and current flowing is in loop no 1
)
Similarly for loop no. 2 currents flowing is resistor R3 it should be )R3
Consider loop no. 1 apply KVL
- …….①
-
Consider loop no. 2 apply KVL
-…….②
-
After solving equation ① and ② we will get branch current and
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
B.
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
Superposition Theorem:
- This is only applicable to circuits with linear elements.
- 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:
- Network A should contain linear elements.
- Network A can have independent and dependent current and voltage source.
- If network A has dependent source than controlling parameter must lie in network A itself.
- Network A should not have any source coupling and magnetic coupling.
- For network B:
- It can have linear and non linear elements.
- It can have dependent and independent voltage and current sources.
- It should not have any source and magnetic coupling with network A.
Method for finding Rth :
Firstly, open circuit terminal A and B.
- If network is operating with only independent sources:
- Make all sources zero in network A.
- Find out the equivalent resistance across terminal A and B.
2. If network A is operating with independent and dependent sources:
- Make all independent sources zero in network A.
- 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:
- Isc =
- 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
- 16Ω
- Ω
- 60Ω
- 20Ω
Solution
Therefore,
Reference
1 D.P Kothari and I.J Nagrath “Basic electrical engineering” Tata Mcgraw Hill,2010
2 D.C.Kulshtreshtha Basic electrical engineering” Tata Mcgraw Hill,2009
3 E.Hughes “Electrical and Electronics Technology” Pearson,2010
4 V.D.Toro “Electrical Engineering Fundamentals” Prentice Hall India, 1989