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12 ^4
10 V 2
B
RTH = 12||6 + 4 = 8
Step 3. To find Vx
After having found the Thevenin circuit, we connect this to the load in order to find VX.
VTH
RTH A
VX
_
V X (^^10 )(^^2 )^ 2 V
4. NORTON’S THEOREM :
Statement : Any two terminal linear active network (containing independent voltage and current sources), may be replaced by a constant current source IN in parallel with a resistance RN, where IN is the current flowing through a short circuit placed across the terminals and RN is the equivalent resistance of the network as seen from the two terminals with all sources replaced by their internal resistance.
10 A
20 ^40
50 V 60
SS
Example : Find the Norton equivalent circuit to the left of terminals A-B for the network shown below. Connect the Norton equivalent circuit to the load and find the current in the 50 resistor.
10 A
20 ^40 ^ A
+ _ 50 V (^60) 50 Solution: (^) B
Fig. Circuit to find INORTON
ISS 10.7 A
It can also be shown that by deactivating the sources,We find the resistance looking into terminals A-B is RN ^^55 RN and RTH will always be the same value for a given circuit. The Norton equivalent circuit tied to the load is shown below.
10.7 A (^) 50
SINGLE PHASE A.C CIRCUIT
Single phase EMF generation:
Alternating voltage may be generated
- By rotating a coil in a magnetic field
- By rotating a magnetic field within a stationary coil
The value of voltage generated depends upon
- No. of turns in the coil 2) field strength 3) speed
Equation of alternating voltage and current
N= No. of turns in a coil
m= Maximum flux when coil coincides with X-axis
= angular speed (rad/sec) = 2f
At =t, = flux component to the plane =m cos t
According to the Faraday‘s law of electromagnetic induction,
dφ d e=-N =-N φ (^) m C osωt = ω N φ (^) m Sinωt ........... (1 ) dt dt
Now, e is maximum value of Em , when Sin = Sin 90 = 1.
i.e Em = Nm ............................................................... (2)
From Eqn^ (1) & (2), e = Em Sin t volt
Now, current (i) at any time in the coil is proportional to the induced emf (e) in the coil. Hence, i = Im Sin t amp
A.C terms :
Cycle:- A complete set of positive and negative values of an alternating quantity is known as cycle.
Time period: The time taken by an alternating quantity to complete one cycle is called time T. Frequency: It is the number of cycles that occur in one second. f = 1/T f = PN/120 where, P= No. of poles, N= Speed in rpm Waveform: A curve which shows the variation of voltage and current w.r.t time or rotation.
Root mean Square (RMS) or effective or virtual value of A.C :-
i 2 +i 2 +.....+i 2 Irms = 1 2 n^ =Square root of the mean of square of the instantaneous currents n
It is the square root of the average values of square of the alternating quantity over a time period. T (^2) I (^) rm s = (^) i (ω t)d (^) ω t (^) 0 Average Value (or mean value) :
It is the arithmetic sum of all the instantaneous values divided by the number of values used to obtain the sum
I a v =
i 1 + i 2 +....... + i n
n
I a v =
T
T 0 i^ ω^ t^ d^ ω t^
Form factor (Kf) :- is the ratio of rms value to average value of an alternating quantity. (Kf = Irms/Iav)
Peak factor (Ka) or crest factor:- is the ratio of peak (or maximum) value to the rms value of alternating quantity (Ka = Imax/ Irms)
Example : An alternating current varying sinusoidally with a frequency of 50 Hz has an RMS value of 20 A. Write down the equation for the instantaneous value and find this value a) 0.0025 sec b) 0.0125 sec after passing through a positive maximum value. At what time, measured from a positive maximum value, will the instantaneous current be 14.14 A?
I (^) m = 2 0 = 2 8 .2 A Ans: (^) ω = 2 π × 5 0 = 1 0 0 π rad /s
The equation of the sinusoidal current wave with reference to point O as zero time point is
i= 28.2 sin 100t Ampere
Since time values are given from point A where voltage has positive and maximum value, the equation may itself be referred to point A. In this case, equation becomes
i= 28.2 cos 100t
i) When t= 0.0025 second i = 28.2 cos 100 X 0.0025 ................. angle in radian = 28.2 cos 100 X 180 X 0.0025 ..................angle in degrees = 28.2 cos 45 = 20 A ................................ point B
ii) When t = 0.0125 sec I =28.2 cos 100 X 180 X 0. = 28.2 cos 225 = 28.2 X (-1/2) = -20 A ...........................................point C
A.C through pure ohmic resistance only
v = iR or i = v = vm^ sinωt^ in phase
R R
A.C through pure inductance only
v= L
di
= Vm sinω t
dt
i=
Vm
L ^
sinω t
i= - Vm
ω L
cosω t
i= I^ sin^ ^ ω^ t-^ π^
^ (current lags by 90^
m 2
ω L= 2 πfL= X L = inductive reactance(in Ω )
A.C through pure Capacitance only
i= C d v = C
d V^ S in^ ω^ t^ d t d t m = ω C V (^) m c o sω t = ω C V s in ^ ω t+ π^ ^ = V^ m^ s in ^ ω t+ π^ m 2 1 2 ω C
= I^ s in^ ^ ω^ t+^ π^ ^ ( c u rre n t le a d s b y 9 0 °) m 2 1 ω C
= X C =
2 π f C
= c a p a c i t iv e re a c ta n c e (^) in (^)
‘j’ operator: j is a operator which rotates a vector by 90 in anticlockwise direction
j^2 = -1 ;j=
Note: ‗i‘ is used for current hence ‗j‘ is used to avoid confusion Mathematical representation of vectors :
1. Rectangular or Cartesian form :- V
2. Polar form : V^ ^ V ^
a jb
3. Trignometrical form : V
V (^) cos j sin (^)
4. Exponential form : V^ ^ V e^ ^ j^ - 1
R=Resistance of the network in ohm. X=Reactance of the electrical network in ohm.
Admittance:
In electrical engineering, admittance is a measure of how easily a
circuit or device will allow a current to flow. It is defined as the inverse of
impedance. The SI unit of admittance is the siemens (symbol S).
Admittance is defined as:
Y = 1/Z
Where
Y is the admittance, measured in siemens Z is the impedance, measured in ohms
The synonymous unit mho, and the symbol ℧ (an upside-down uppercase omega
Ω), are also in common use.
Resistance is a measure of the opposition of a circuit to the flow of a steady
current, while impedance takes into account not only the resistance but also
dynamic effects (known as reactance). Likewise, admittance is not only a measure
of the ease with which a steady current can flow, but also the dynamic effects of
the material's susceptance to polarization:
Y = G + j B
Where
Y is the admittance, measured in siemens. G is the conductance, measured in siemens. B is the susceptance, measured in siemens.
AC Equivalent Circuits :
- Impedances in series add together to give the equivalent impedance while the
admittance in parallel add together to give the equivalent admittance.
- Impedances in parallel gives equivalent impedance by reciprocating the
reciprocal sum of the impedances and to obtain the equivalent admittance in series
same procedure has to be followed.
Instantaneous and Average Power
The most general expressions for the voltage and current delivered to an arbitrary
load are as follows:
v(t) = V cos(ωt − θV )
i(t) = I cos(ωt − θI )
