circuit gyrator provides ——– phase shift in forward direction. Gyrator definition and meaning | Gyrator coefficient depends upon what factors ?
Basic Concepts
Electrical Network Network analysis is any structured technique used to mathematically analye a circuit. A physical electrical network or electrical circuit is a system of interconnected components. the word components include source of energy such as voltage sources or current sources, electrical elements such as resistors, inductors ans capacitors.
Electrical Network Components
Resistor
Linear and bilateral element
V(t) = RI (T)
I (T) = V (T)/R (in time domain)
V(S) = RI (S)/R (in s-domain s = 0 = j00)
(complex frequency)
V = RI (in frequency domain)
I = V/R (for sinusoidal excitation)
Inductor
V(T) = L d/dt I (T)
I(T) = 1/L V (T) dt (in time domain)
V (S) = sL I (S)
I(S) = 1/sL V(S)
ZL = sL = Impedance
V = J00 L I
I = 1/J00L V (in frequency domain)
(for sinusoidal excitation)
ZL = J00L and YL = 1/JOOL
1. The representation of any element in the s-domain, automatically includes the initial condition in the given element.
2. any differential or integral equation in the time domain is transformed to s-domain and therefore, is converted to linear equations.
Therefore, the manipulation of the equations become simpler.
Capacitor
I (T) = C dv(t)/dt (in time domin)
V (T) = 1/C I (T) dt
I (S) = sCV (s)
V (s) = 1/C (s) = I (s) (in s-domain)
ZC = 1/sC and YC = sC
I = J00C V (in frequency domain)
V = 1/J00C I
ZC = 1/J00C
YC = J00C
Transformer
V1/V2 = N1/N2
I1/I2 = N2/N1
Gyrator
RO = Gyrator coefficient depends upon
(a) op-amp parameters
(b) R, C values
V1 = R0I2
V2 = – R0I1
. A gyrator is a four-terminal or a 2-port device.
. The coefficient of the gyrator depends upon the op-amp parameters and externally connected R,C values.
. This device is used to simulate equivalent value of inductance L.
. The device shows an impedance inversion therefore, for a inductive load impedance, the input impedance is capacitive and vice-versa.
Dependent Sources
. The handling of independent and dependent voltage sources or current sources remain exactly same except in
(a) Thevenin’s and norton’s model for the calculation of equivalent impedance.
(b) Superposition therorem.
2. All ideal independent voltage sources are short-circuited.
All ideal independent current sources are open-circuited.
3. All dependent voltage and curent sources remain as they are
Solution of any Network
Using KVL and KCL Analysis
Number of KCL equations = (n – 1) = (3 – 1) = 2
number of KCL equations = b – (n – 1) = 5 – (3 – 1) = 3
voltage drop = positive
voltage rise = negative KVL
= negative
positive KCL
Kirchhoff’s Voltage Law – 1
– 5 + 1 x I1 + 2 (I1 + I2) = 0
3I1 + 2I2 + 0I3 = 5 ………..(1)
Kirchhoff’s Voltage Law – 2
4(I2 + I3) + 3I2 + 2(I1 + I2) = 0
2I1 + 9I2 + 4I3 = 0………………..(2)
Kirchhoff’s Voltage Law-3
10 + 5I3 + 4(I2 + I3) = 0
0I2 + 4I2 + 9I3 = – 10 ………….(3)
[Z] = Z11 Z12 Z13
Z21 Z22 Z23
Z31 Z32 Z33
Z12 = Z21
Z13 = Z31 Z12 = Z21
Z23 = Z32 V1/I2 = V2/I1
(network is reciprocal)
(reciprocity theorem is applicable)
Kirchhoff’s Current law-1
+ (VA – 5/1 + VA – VB/3 + VA – 0/2 = 0
VA (1 + 1/3 + 1/2) + VB (-1/3) = 5
VA (11/6) + VB (-1/3) = 5 ……………(1)
Kirchhoff’s Current law-2
VB – VA/3 + VB – 0/4 + VB – 10/5 = 10
VA (-1/3) + VB (1/3 + 1/4 + 1/5) = 2
VA (-1/3) + VB (47/60) = 2 ………….(2)
[11/6 – 1/3/-1/3 47/60] [VA/VB] = [5/2]
[Y11 Y12/Y21 Y22] [VA/VB] = [I1/I2]
Y12 = Y21 (Network is reciprocal)
[Z] [I] = [V] ………….KVL
[Y] [V] = [I] ………….KCL
Example 1 Find the voltage across R in the network given in figure by mesh analysis.
Sol. here, b = 8, s = 1, n = 6 and number of mesh equations,
m = b – n + s
m = b – (n – 1)
m = 8 – (6 – 1)
m = 3
in loop-I, by kirchhoff’s voltage law (KVL)’
(10 + 2)I1 + (-2)I2 + (0)I3 = 5
(12)I1 + (-2)I2 + (0)I3 = 5 …………..(1)
In loop – II
(-2)I1 + (10 + 2 + 20 + 2)I2 + (-2)I3 = 0
(-2)I1 + (34)I2 + (-2)I3 = 0……………….(2)
In loop – III
(0)I1 + (-2)I2 + (2 + 10)I3 = 10
(0)I1 + (-2)I2 + (12)I3 = 10 …………..(3)
Rearranging the above equations, we can write the mesh equations in matrix form as
[12 -2 0/-2 34 -2/0 -2 12] [I1/I2/I3] = [5/0/10]
The voltage across R is
VR = (I2 – I1)R
I2 = 360/4800 = 3/40 A
[12 5 0/-2 0 -2/0 10 12] = 360
[12 -2 0/-2 34 -2/0 -2 12]= 4800
I3 = 4060/4800 = 203/240 A
[12 -2 5/-2 34 0/0 -2 10]
= 4060
VR = (I2 – I3) R
VR = 1.54 V
Network Equations for RLC network
Using kirchhoff’s laws, we can write loop equations. referring to fig. 8 using KVL, we have,
RI + L dI/dt + 1/C [I. dt = V (t)
Taking derivatives of both sodes, this equation can be written as
L = d2I/dt2 + R dI/dt + 1/C I = dV (t)/dt
Using KVL in figure, we have the mesh equations as
Loop I,
R1I1 + 1/C1 (I1 – I2) dt = V (t)
loop II,
1/C1 (I2 – I1) dt + L1 dI2/dt + 1/C2 (I2 – I3) dt = 0
loop = III
I/C2 (I3 – I2) + L2 dI3/dt + I3R2 = 0
Using KCL, the node equations of figure may be written as
Node 1, 1/R1 V1 + C d/dt (V1 – V2) = I1 (t)
Node 2, C d/dt (V2 – V1) + 1/R2 V2 = 0
Example 2. For the network shown in figure, determine the node voltages.
Sol. the nodal equations can be written directly in the matrix form after transformation (taking node 3 as datum node), as
[1/5 + 1/j2 + 1/4 – 1/4 /-1/4 1/4 + 1/-j2 + 1/2] [V1/V2] = [5/2 <00/5/2 <900] = [1/J2.5]
V1 = 1 |1/J2.5 -0.25/0.75 + J0.5|
V1 = 2.48 < 72.10
V2 = 3.44 < 26.470
|0.45 – J0.5 /-0.25 0.75 + J0.5|
= 0.545 < – 15.80
Intro Exercise-1
1. The energy required to move 120 C through 3 V is
(a) 25 mj
(b) 360 j
(c) 40 j
(d) 2.78 mj
2. Find the current I flowing in the given figure.
(a) 1 A
(b) 2 A
(c) 3 A
(d) 4 A
3. The dependent current source shown in figure
(a) delivers 80 w
(b) absorbs 80 w
(c) delivers 40 w
(d) absorbs 40 w
4. for a given voltage, four heating colis will produce maximum heat, when connected
(a) all in parallel
(b) all in series
(c) with two parallel pairs in series
(d) one pair in parallel with the other two in series
5. In the circuit of the figure, the value of the voltage source E is
(a) – 16 V
(b) 4 V
(c) – 6 V
(d) 16 V
6. In the circuit of the figure a charge of 600 c is delivered to the 100 v source in 1 min. the value of V1 must be
(a) 240 v
(b) 120 v
(c) 60 v
(d) 30 v
7. For the circuit shown in figure, the value of voltage V0 is
(a) 10 v
(b) 15 v
(c) 20 v
(d) none of these
8. The energy stored in the magnetic field at a solenoid 30 cm long and 3 cm diameter wound with 1000 turns of wire carrying a current of 10 A is
(a) 0.015 j
(b) 0.15 j
(c) 0.5 j
(d) 1.15 j
9. All resistances in the circuit in the figure are at R ohm each. the switch is initially opened. what happens to the intensity of lamp when the switch is closed?
(a) Intensity of lamp increases
(b) intensity of lamp decreases
(c) intensity of lamp remains same
(d) answer depends on the value at R
10. The time constant of the network shown in the figure is
(a) 2 RC
(b) 3 RC
(c) RC/2
(d) 2 RC/3
11. At a certain current, the energy stored in iron cored coil is 1000 j and its copper loss is 2000 w. the time constant (in second) of the coil is
(a) 0.25
(b) 0.5
(c) 1.0
(d) 2.0
12. The current waveform in a pure resistor at 10 is shown in figure. power dissipated in the resistor is
(a) 7.29 w
(b) 52.4 w
(c) 135 w
(d) 270 w
13. In the figure below the, current of 1 A flows through the resistance of
(a) 4
(b) 20
(c) 30
(d) 12
14. Find the value of VX in the given figure.
(a) 37.5 v
(b) 5 v
(c) 32.5 v
(d) 100 v
15. Find the value of CEQ in the given figure.
(a) 3.5 uf
(b) 1.2 uf
(c) 2.4 uf
(d) 4.05 uf
16. In the circuit of figure below, bulb a uses 48 w when lit, bulb b uses 22 w when lit and bulb c uses 14.4 w when lit. the additional bulbs in parallel to this circuit, that would be required to below the fuse, are
(a) 4
(b) 5
(c) 6
(d) 7
17. Find currents I1,I2 and I3 in the given figure.
(a) 1.38 a, 1.1 a, 5.52 a
(b) 1.38 a, 5.5 a, 1.1 a
(c) 1.38 a, 0.9 a, 6.38 a
(d) 1.1 a, 5.52 a, 1.38 a
18. The network shown in the figure, effedtive resistance faced by the voltage source, is
(a) 4
(b) 3
(c) 2
(d) 1
19. In the circuit of figure, the power absorbed by the load RL is
(a) 2 w
(b) 4 w
(c) 6 w
(d) 8 w
20. In the circuit of figure dependent source
(a) supplies 16 w
(b) absorbs 16 w
(c) supplies 32 w
(d) absorbs 32 w
1. (b)
W = QV
= 120 x 3
= 360 J
2. (a)
3. (a)
applying KVL,
– 20 + 5i + 5 ( I + V1/5) = 0
10i = 20 – V1
V1 = 20 V
10i = 20 – 20
I = 0
Only dependent source acts
V1/5 = 20/5 = 4 A
Power delivered = I2R = (4)2 x 5
= 16 x 5 = 80 w
4. (a)
REQ = R||R||R||R||
= R/4
Maximum heat produced
P = V2/REQ
= V2/(R/4)
= 4V2/R
So, for a given parallel network REQ is minimum and hence, maximum heat produced because
P OC 1/REQ
5. (a)
going from 10 v to 0 v
10 – 0 = – 5 – E – 1
10 = – 6 – E
E = – 16 V
6. (a)
we know that
I = dQ/dt = 600/60 = 10 A
charge is delivered to 100 V source, hence the current must be anti-clockwise,
by applying KVL to the circuit,
V1 + 60 – 100 = 10 x 20
= V1 = 240 V
7. (b)
voltage is constant because of 15 v source.
V0 = 15 V
L = N2U0A/I
= 106 x 4 x 10-7x 4 x (9 x 10-4)/0.3
= 92 x 10-5/0.3
energy E = 1/2 LI2
= 1/2 x 92 x 10-5/0.3 x (10)2
= 0.148 or 0.15 J
9. (c)
When switch is closed intensity of lamps remains same.
10. (c)
first replace the voltage source by their internal impedance
REQ = R x 2R /3R = 2R/3
time constant
= REQ x C = 2/3 RC
11. (c)
time constant t = L/R = I2L/I2R
= 2 x 1000/2000 = 1
= 1 s
12. (d)
power = I2R
= I2MUS xR
I2MUS = 1/3 (9/3 t)2dt
= [1/3 x 9 x t3/3] = 27 A2
Power P = 27 x 10
= 270 W
13. (d)
there are three parallel resistance banks which are connected in series
REQ = 20||30
= 20 x 30/20 + 30
= 12
REQ2 = 6||4||12
= 2
REQ3 = 10||15
= 10 x 15 /25 = 6
REQ = REQ1 + REQ2 + REQ3 = 12 + 2 + 6 = 20
current drawn from supply of 120 v
= 120/I = 120/20 = 6 A
Current through 20 resistor
= 6 x 30/20 + 30 = 3.6 A
Current through 30 resistor
= 6 x 30/20 + 30 = 3.6 A
current through 30 resistor
= 6 – 3.6 = 2.4 A
REQ2 = 6||4||2
= 20
voltage across REQ2 = 6 x 2 = 12 v
current through 6 resistor = 12/6 = 2 A
current through 4 resistor = 12/4 = 3 A
current through 12 resistor = 12/12 = 1 A
14. (b)
current flowing through 5 resistor is 1 A
VX = V5 = 1 x 5
VX = 5 V
15. (d)
1 uf is in parallel with 2.5 uf and this combination is in series with 1.5 uf.
C1 = 1.5 (2.5 + 1)/1.5 + 2.5 + 1 = 5.25/5 = 1.05
C1 is in parallel with 3 uf
CEQ = 1.05 + 3 = 4.05 uf
16. (a)
IA = 48/12 = 4 A
IB = 22/12 = 1.8 A
IC = 14.4/12 = 1.2 A
Current required below fuse = 20 A
so, additional bulbs must draw currents
= 20 – (4 + 1.8 + 1.2)
= 20 – 7 = 13 A
number of additional bulbs = 13/3 = 4.3
so, 4 additional bulbs are required.
17. (a)
We cannot apply current divider rule because there are three resistors in parallel.
I1 = 4 (5/6)/4 + (5/6) x 8
= 1.38 A
I4 = 4/4 + 5/6 x 8
I4 goes into parallel combinations of 5 and 1
I2 = 4 x 1.38/5 = 1.1 A
I3 = 4 x 1.38/1 = 5.52 A
18. (b)
Applying KVL, -V + 3I/4R = 0
V = 3I/4R = 3I/4 x R
V/I = 3/4 R
V/I = 3/4 x 4
V/I = 3 = R’
19. (d)
I1 = V/R = 1/1 = 1 A
Current through RL = 2I1 = 2 A
Power P = I2R = (2)2 x 2
= 8 W
20. (d)
Power P = VI
= (2IX) x IX = 2I2X
= IX = 4 A
P = 2 (4)2
32 W (absorb)