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.
Ohms Law: It states that the current through the conductor between two points is directly proportional to the voltage across the two points
V α I
V= IR
V = voltage across terminals in volts
R = resistance in ohms
I= current in Amperes
The algebraic sum of currents meeting at a junction or node in an 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
i.e. 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.
i.e. ∑ 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 i.e. 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 do 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
And Norton Theorems.
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 nonlinear 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,
13Isc=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
FOR RL Circuit
Fig: Series RL circuit
After switch is closed applying KVL
=0
This is first order homogeneous differential equation so
dt
Integrating both sides
Ln i= t+K
Taking antilog of both sides
i=k
At t=0
i(0)==I0
=ke0
The particular solution is given as
i= for t≥0
= for t<0
FOR RC Circuit
Fig: Series RC circuit
=V
For t>0 applying KVL
Ri(t)++V=0
Hence general solution of above equation is calculated same as for RL circuit
i=k
i(0)=
Hence, particular solution for network is given as
i= for t for t≥0
= for t<0
Time Constant
Time constant for RL circuit
From above section for RL circuit at t≥0
i=
i=I0
I0=
Time taken for current to drop from unity to zero is called as time constant T.
sec
FOR RC Circuit
It can be calculated in the same manner as for series RL circuit
The time constant is given as
T=RC
Average Value:
The arithmetic means of all the value over complete one cycle is called as average value
=
For the derivation we are considering only hall cycle.
Thus varies from 0 to ᴫ
i = Im Sin
Solving
We get
Similarly, Vavg=
The average value of sinusoid ally varying alternating current is 0.636 times maximum value of alternating current.
RMS value: Root mean square value
The RMS value of AC current is equal to the steady state DC current that required to produce the same amount of heat produced by ac current provided that resistance and time for which these currents flows are identical.
I rms =
Direction for RMS value:
Instantaneous current equation is given by
i = Im Sin
But
I rms =
=
=
=
Solving
=
=
Similar we can derive
V rms= or 0.707 Vm
the RMS value of sinusoidally alternating current is 0.707 times the maximum value of alternating current.
Real Power: [P]
It is nothing but the actual power being used in a circuit.
P= = I2R Watts
Reactive Power: [Q]
It is the function of reactance in the circuit X. Mainly reactive loads are inductor and capacitors. These elements dissipate zero power. These element shows that they dissipate power. This is called as reactive power.
Q= = I2X VAR (voltAmpereReactive)
Apparent Power: [S]
It is the product of a circuit voltage and current without reference to phase angle. It is the combination of both reactive and real power.
S= = I2Z VA (voltAmpere)
Power factor (P.F.)
It is the cosine of angle between voltage and current
If Ɵis –ve or lagging (I lags V) then lagging P.F.
If Ɵ is +ve or leading (I leads V) then leading P.F.
If Ɵ is 0 or in phase (I and V in phase) then unity P.F.
Reactance
 Inductive Reactance (XL)
It is opposition to the flow of an AC current offered by inductor.
XL = ω L But ω = 2 ᴫ F
XL = 2 ᴫ F L
It is measured in ohm
XL∝FInductor blocks AC supply and passes dc supply zero
2. Capacitive Reactance (Xc)
It is opposition to the flow of ac current offered by capacitor
Xc =
Measured in ohm
Capacitor offers infinite opposition to dc supply
Impedance (Z)
The ac circuit is to always pure R pure L and pure C it well attains the combination of these elements. “The combination of R1 XL and XC is defined and called as impedance represented as
Z = R +i X
Ø = 0
only magnitude
R = Resistance, i = denoted complex variable, X =Reactance XL or Xc
Polar Form
Z = L I
Where =
Measured in ohm
Ac circuit containing pure resisting
Consider Circuit Consisting pure resistance connected across ac voltage source
V = Vm Sin ωt ①
According to ohm’s law i = =
But Im =
②
Phases diagram
From ① and ②phase or represents RMD value.
Power P = V. i
Equation P = Vm sin ω t Im sin ω t
P = Vm Im Sin2 ω t
P = 
Constant fluctuating power if we integrate it becomes zero
Average power
Pavg =
Pavg =
Pavg = Vrms Irms
Series RLC circuit
The voltage across inductor = jIXL
Voltage across capacitance = jIXC
Net voltage across L and C = jI(XLXC) = j(ELEC)
Voltage drop across R= IR=ER
Applied voltage E= IR+jI(XLXC)
E=I
Z=R+j(XL+XC)
Z=
Φ= tan1
I=
Cos Φ= R/Z
Active Power= EI CosΦ
Reactive Power = EI SinΦs
Definition: it is defined as the phenomenon which takes place in the series or parallel RLC circuit which leads to unity power factor
Voltage and current in R – L  C ckt. Are in phase with each other
Resonance is used in many communicate circuit such as radio receiver.
Resonance in series RLC series resonance in parallel RLC anti resonance / parallel resonance.
 Condition for resonance XL = XC
 Resonant frequency (Fr): for given values of RLC the inductive reactance XL become exactly equal to the capacitive reactance Xc only at one particular frequency. This frequency is called as resonant frequency and denoted by (fr)
 Expression for resonant frequency(fr): we know that XL = 2ƛ FL  Inductive reactance
Xc =  capacitive reactance
At a particular frequency ȴ = fr, the Inductive and capacitive reactance are exactly equal
XL = XC ……at ȴ = fr
I.e. L =
fr2 =
fr = H2
And = wr = rad/sec
Quality factor / Q factor
The quality of resonance circuit is measured in terms of efficiency of L and C to stare energy and the efficiency of L and C to store energy as measured in terms of a factor called quality factor or Q factor it is expressed as
Q = and Q =
The sharpness of tuning of RLC series circuit or its selectivity is measured by value of Q. As the value of Q increases, sharpness of the curve also increases and the selectivity increases.
Bandwidth (BW) = f2 = b1
and are the frequency at which the power delivered to the resistor is reduced to 50% of the power delivered to it at resonance these frequencies are called as half power frequency
Bw = fr/Q
Resonance in Parallel circuit:
When a coil is in parallel with a capacitor, as shown below. The circuit is said to be in resonance.
The resonant frequency for above circuit is fr = Hz
The current at resonance is I=
The value L/RC is known as dynamic impedance.
The current at resonance is minimum. The circuits admittance must be at its minimum and one of the characteristics of a parallel resonance circuit is that admittance is very low limiting the circuits current. Unlike the series resonance circuit, the resistor in a parallel resonance circuit has a damping effect on the circuit bandwidth making the circuit less selective.
Also, since the circuit current is constant for any value of impedance, Z, the voltage across a parallel resonance circuit will have the same shape as the total impedance and for a parallel circuit the voltage waveform is generally taken from across the capacitor.
Bandwidth and selectivity:
and are the frequency at which the power delivered to the resistor is reduced to 50% of the power delivered to it at resonance these frequencies are called as half power frequency
Bw = fr/Q
Q = = fCR = R
Resonant Frequency:
The resonant frequency for parallel resonant circuit is given as
fR=
Where L= inductance of the coil
C = is the capacitance
Rs = Resistive value of coil.
Que) A coil takes a current of 6A when connected to 24V dc supply. To obtain the same current with 50HZ ac, the voltage required was 30V. Calculate inductance and p.f of coil?
Sol: The coil will offer only resistance to dc voltage and impedance to ac voltage
R =24/6 = 4ohm
Z= 30/6 = 5ohm
XL =
= 3ohm
Cosφ = = 4/5 = 0.8 lagging
Que) The potential difference measured across a coil is 4.5V, when it carries a dc current of 8A. The same coil when carries ac current of 8A at 25Hz, the potential difference is 24V. Find current and power when supplied by 50V,50Hz supply?
Sol: R=V/I= 4.5/8 = 0.56ohm
At 25Hz, Z= V/I=24/8 =3ohms
XL =
= 2.93ohm
XL = 2fL = 2x 25x L = 2.93
L=0.0187ohm
At 50Hz
XL = 2x3 =6ohm
Z = = 5.97ohm
I= 50/5.97 = 8.37A
Power = I2R = 39.28W
Que) A coil having inductance of 50mH an resistance 10ohmis connected in series with a 25F capacitor across a 200V ac supply. Calculate resonant frequency and current flowing at resonance?
Sol: f0= = 142.3Hz
I0 = V/R = 200/10 = 20A
Que) A 15mH inductor is in series with a parallel combination of 80ohm resistor and 20F capacitor. If the angular frequency of the applied voltage is 1000rad/s find admittance?
Sol: XL = 2fL = 1000x15x103 = 15ohm
XL = 1/C = 50ohm
Impedance of parallel combination Z = 80j50 = 22.5j36
Total impedance = j15+22.5j36 = 22.5j21
Admittance Y= 1/Z = 0.023j0.022 siemens
Que) A circuit connected to a 100V, 50 Hz supply takes 0.8A at a p.f of 0.3 lagging. Calculate the resistance and inductance of the circuit when connected in series and parallel?
Sol: For series Z =100/0.8 = 125ohm
Cosφ =
R = 0.3 x 125 = 37.5ohm
XL = = 119.2ohm
XL = 2fL = 2x 50x L
119.2 = 2x 50x L
L= 0.38H
For parallel:
Active component of current = 0.8 cosφ = 0.3x0.3 = 0.24A
R = 100/0.24 =416.7ohm
Quadrature component of current = 0.8 sinφ = 0.763
XL= 100/0.763 = 131.06ohm
L= 100/0.763x2x50 = 0.417H