for a parallel rlc circuit the incorrect statement among the following is , for parallel rlc circuit which one of the following statements is not correct ? answers given below after questions.
Unit Exercise – 1
1 Mark Questions
- For the two-port network shown below, the short-circuit admittance parameter matrix is
(a) [4 -2/-2 4]
(b) [1 -0.5 /-0.5 1]
(c) [1 0.5 /0.5 1]
(d) [4 2/2 4]
- For parallel RLC circuit, which one of the following statements is not correct?
(a) the bandwidth of the circuit decreases increased
(b) the bandwidth of the circuit remains same if L is increased
(c) at resonance, input impedance is a real quantity
(d) at resonance, the magnitude of input impedance attains its minimum value
- The nodal method of circuit analysis is based on
(a) KVL and ohm’s law
(b) KCL and ohm’s law
(c) KCL and KVL
(d) KCL, KVL and ohm’s law
- The equivalent capacitance for the network shown in the figure is
(a) c/4
(b) 5c/13
(c) 5c/2
(d) 3c
- Twelve 1 resistors are used as edge to form a cube. the resistance between two diagonally opposite corners of the cube is
(a) 5/6
(b) 6/5
(c) 1
(d) none of these
- Twelve 1 H inductors are used as edge to form a cube. the inductance between two diagonally opposite corners of cube is
(a) 5/6 H
(b) 10/6 H
(c) 2 H
(d) 3/2 H
- The equivalent resistance as seen between the terminals (a, b) is
(a) 2
(b) 4
(c) 1
(d) not possible
- A fully charged mobile phone with a 12 V battery is good for a 10 min. talk-time. Assume that during the talk-time, the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. how much energy does the battery deliver during this talk-time?
(a) 220 j
(b) 12 kj
(c) 13.2 kj
(d) 14.4 kj
- In th interconnection of ideal sources shown in the figure, it is known that the 60 v source is absorbing power.
Which of the following can be the value of the current source?
(a) 10 A
(b) 13 A
(c) 15 A
(d) 18 A
- If the transfer function of the following network is VO(s)/V1(s) = 1/2 + sCR
The value of the load resistance R is
(a) R/4
(b) R/2
(c) R
(d) 2R
- Given that F(s) is the one-sided laplace transform of f(t), the laplace transform of is
(a) sf(s) – f(0)
(b) 1/s F(s)
(c) F d
(d) 1/s [F(s) – f(0)]
- The network element and V-I characteristics are shown in fig. (a) and fig. (b). the element is
(a) non-linear, active and bilateral
(b) linear, passive and bilateral
(c) non-linear, passive and non-bilateral
(d) non-linear, active and non-bilateral
- The current having the waveform shown in figure is flowing in a resistance of 1. the average power is
(a) 1000 W
(b) 1000/2 W
(c) 1000/3 W
(d) 1000/4 W
- The current l(t) in the conductor A is as shown in figure. the charge that is transported by this current through surface SO of the conductor during the time interval 0 < t < 6 s is
(a) 7.5 C
(b) 8.5 C
(c) 5.5 C
(d) zero
- Current through and voltage across a device are given in figure. energy (approximate) absorbed by the device in the interval 0 < t < 2 s is
(a) 8.5 j
(b) 10 j
(c) 11.4 j
(d) 5 j
- Given the current through R is l = 1 A. the value of R is
(a) 0
(b) 2
(c) 4
(d) 8
- The equivalent inductance of the network is
(a) L1 + L2 – 2 M
(b) L1 + L2 + 2 M
(c) L1 + L2 – M
(d) L1 + L2
- Which one of the following networks is the Y-equivalent of circuit shown in figure?
- Consider the circuit graph shown in figure. each branch of circuit graph represents a circuit element. the value of voltage V1 is
(a) -30 V
(b) 25 V
(c) -20 V
(d) 15 V
- In the following circuit, the switch S is closed at t = 0. the rate of change of current dl/dt (0+) is given
(a) 0
(b) RSIS/L
(c) (R + RS)IS/L
(d) oo
- In the following graph, the number of trees (p) and the number of cut-sets (Q) are
(a) p = 2, q = 2
(b) p = 2, q = 6
(c) p = 4, q = 6
(d) p = 4, q = 10
- The pole- zero plot given below corresponds to a
(a) low-pass filter
(b) high-pass filter
(c) band-pass filter
(d) notch filter
- Consider the following graphs :
Non-planar graph(s) is/are
(a) 1 and 3
(b) 4 only
(c) 3 only
(d) 3 and 4
- Consider the following circuits :
The planar circuits are
(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 4 and 1
- A tree of the graph shown below is
(a) a d e h
(b) ac f h
(c) a f h g
(d) ae f g
- For the graph shown below correct set is
Node branch twigs link
(a) 4 6 4 2
(b) 4 6 3 3
(c) 5 6 4 2
(d) 5 5 4 1
- In the following circuit, the voltage VA is
(a) 4.33 V
(b) 4.09 V
(c) 8.67 V
(d) 8.18 V
- In the following circuit, the voltage V2 is
(a) 0.5 V
(b) 1.0 V
(c) 1.5 V
(d) 2.0 V
- In the following circuit, the voltage V1 is
(a) 120 V
(b) -120 V
(c) 90 V
(d) -90 V
- In the following circuit, the voltage V1 is
(a) 0.4 VS
(b) 1.5 VS
(c) 0.67 VS
(d) 2.5 VS
- In the following circuit the voltage VA is
(a) -11 V
(b) 11 V
(c) 3 V
(d) -3 V
- An independent voltage source in series with an impedance ZS = RS = JXS delivers a maximum average power to a load impedance ZL when
(a) ZL = RS + JXS
(b) ZL = RS
(c) ZL = JXS
(d) ZL = RS = JXS
- The RC circuit shown in the figure is
(a) a low-pass filter
(b) a high-pass filter
(c) a band-pass filter
(d) a band-reject filter
- The condition on R,L and C such that step response y(t) in figure has no oscillations, is
(a) R > 1/2 L/C
(b) R > L/C
(c) R > 2 L/C
(d) R = 1/LC
- The ABCD-parameters of an ideal n:1 transformer shown in figure are [n 0 0 x] the value of x will be a
(a) n
(b) 1/n
(c) n2
(d) 1/n2
- In a series RLC circuit R = 2 K, L = 1 H and C = 1/400 uf. the resonant frequency is
(a) 2 x 104 hz
(b) 1 x 104 hz
(c) 104 hz
(d) 2 x 104 hz
- The maximum power that can be transferred to the load resistor RL from the voltage source in figure is
(a) 1 W
(b) 10 W
(c) 0.25 W
(d) 0.5 W
- All the resistances in the figure are 1 each. the value of I will be
(a) 1/15 A
(b) 2/15 A
(c) 4/15 A
(d) 8/15 A
- The time constant of the network shown in the figure is
(a) 4RC
(b) 3RC
(c) 4RC
(d) 2RC/3
- 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
- In the following circuit, the values of VTH and RTH are
(a) – 2 V, 6/5
(b) 2V, 5/6
(c) 1 V, 5/6
(d) -1 V, 6/5
- A simple equivalent circuit of the two-terminal network shown in figure is
- In the following circuit, the values of IN and RN are
(a) 3 A, 10/3
(b) 10 A, 4
(c) 1.5 A, 6
(d) 1.5 A, a
- In the following circuit, the values of VTH and RTH are
(a) 2 V, 4
(b) 4 V, 4
(c) 4 V, 5
(d) 2 V, 5
- In the following circuit, the values of IN and RN are
(a) 4 A, 3
(b) 2 A, 6
(c) 2 A, 9
(d) 4 A, 2
- For the following circuit, the values of VTH and RTH are
(a) – 100 V, 75
(b) 155 V, 55
(c) 155 V, 37
(d) 145 V, 75
- Consider the network graph shown in figure. which one of the following is not a tree of this graph?
- The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in figure, is
(a) L1 + L2 + M
(b) L1 + L2 – M
(c) L1 + L2 + 2M
(d) L1 + L2 – 2M
- The circuit shown in figure with R = 1/3, L = 1/4 H, C = 3F has input voltage V(t) = sin 2t. the resulting current I(t) is
(a) 5 sin (2t + 53.10)
(b) 5 sin (2t + – 53.10)
(c) 25 sin (2t + 53.10)
(d) 25 sin (2t – 53.10)
- For the RL circuit shown in figure the input voltage v(t) – u(t). the current I(t) is
Answers and solution
Unit Exercise – 1
- (a) Short-circuit admittance parameters are given by
Two-port current in terms of voltage
[I1/I2] = [Y11 Y12/Y21 Y22] [V1/V2]
I1 = Y11 V1 + Y12 V2
I2 = Y22 Y1 + Y22 V2
Y11 = I1/V1|V2 = 0
Y12 = I1/V2|V1 = 0
Y21 = I2/V1|V2 = 0
Y22 = I2/V2|V1 = 0
To calcuate Y11 and Y21
Terminals 2 – 2′ make shorten V2 = 0
Now, circuit is
Y11 = I1/V1|V2 = 0 = 1/0.5110 x 5 = 1/0.25 = 4
V1 = – 0.5 I2
I2/V1 = – 1/0.5 I2
I2/V1 = – 1/0.5 = – 2
Y21 = I2/V1 = – 2
To calcuate Y12 and Y22
Terminals 1 – 1′ make shorten as V1 = 0.
Now, circuit is
V2 = 0.5I1
Y12 = I1/V2|V1 = 0 = 1/-0.5 = – 2
V2 = (0.5110.5)I2 = 0.25I2
Y22 = I2/V2|V1 = 0 = 1/0.25 = 4
Hence, [Y11 Y12/Y21 Y22] = [4 -2/-2 4]
Alternate Method
For general network
The admittance parameters
Y11 = YA + YC
Y12 = – YC
Y21 = – YC
Y22 = YC + YB
Convert impedance to admittance in circuit
Y11 = 2 + 2 = 4
Y12 = – 2
Y21 = – 2
Y22 = 2 + 2 = 4
[Y11 Y12/Y21 Y22] = [4 -2 /-2 4]
- (d)
parallel resonant circuit
Q = 00/s0
so = o0/Q
- OORC = R/0OL = R/R C/L
so = BO = 1/LC x 1L/RL = 1/RC
So = BO = 1/LC x 1/L/RC = 1/RC
O0 = 1/LC
Hence, bandwidth of parallel resonant circuit
BW = 1/RC
i.e., BW decreases when R is increased and remains constant if L is increased.
Impedance curve of parallel resonant circuit At resonance, from figure input impedance is a real quantiy and attains its maximum value.
- (b)
Nodal circuit analysis method is based on KCL and ohm’s law.
- (b)
Circuit for equivalent capacitance by applying the concept of series parallel combination of the capacitance.
1/CEQ = 1/C + 1/C + 1/5C/3
= 1/C + 1/C + 3/5C = 1/C [5 + 5 + 3/5]
CEQ = 5C/13
- (a)
Suppose value of resistance be R each.
Applying KVL by selecting any path between A and B,
VAB = 1/3. R + 1/6 R + 1/3 R
VAB = IR [5/6]
VAB/I = 5R/6
RAB = 5R/6
Where, R = 1 for this problem
so, RAB = 5/6
- (a)
Apply same concept as done in earlier problem.
- (a)
RAB = 1 + (2||RAB)
RAB = 1 + 2.RAB/2 + RAB
RAB = 2 + RAB + 2RAB/2 + RAB
2RAB + (RAB)2 = 2 + RAB + 2RAB
(RAB)2 – RAB – 2 = 0
R2AB – 2RAB + RAB – 2 = 0
RAB (RAB – 2) + 1 (RAB – 2) = 0
(RAB + 1) (RAB – 2) = 0
RAB = – 1,2
RAB = – 1not possible
RAB = 2
- (c)
Energy delivered during talk-time
E = V (t) I (t) dt
I (t) = 2 A = constant = 2 |v(t) dt
2 x shaded area
= 2 x 1/2 x (10 + 12) x 60 x 10
= 13.2 kj
- (a)
Given, 60 V source is absorbing power, it means that current flow from +ve to -ve terminal in 60 V source (as indicated I1 in figure)
I + I1 = 12 A ………….(i)
current source must be have the value less than 12 A to satisfies eq . (1).
- (c)
Impedance of capacitor XC = 1/sC
VO = ILRI ………………….(i)
From current division rule,
IL = I x XC/RL + XC
VO = I x CRL/RL + XC ………..(ii)
Equivalent impedance of circuit
ZEQ = R + (XC||RL)
= R + XCRL/XCRL
We can write
I = VI/ZEQ
Putting value in eq. (ii)
VO (S) = VI (s)/XCRL x XCRL/XC + RL
VO(s)/VI (s) = XCRL/RXC + RRL + XCRL
Dividing by XCRL,
VO(s)/VI (s) = 1/1 + R/XC + R/RL = 1/1 + RsC + R/RL
R = RL
VO/VI (S) = 1/2 + sCR
- (b)
Laplace transform of definite integral
L[f(t) dt] = F(s)/s
laplace transform of indefinite integral
L[ f (t) dt] = F(s)/s + r-1(0+)/s
- (d)
The given figure is non-linear because after 1 V the output becomes constant. any element will be passive if the ratio V|I is positive any time, so the given output shown that the element in passive.
Element will be bilateral if and only if the magnitude (i.e., V or I) is same in both the directions. here, the given element is non-bilateral.
- (c)
Average power = [1/1 (10t)2 dt ] R
[By using formula average power = 1/T |I2 dt]. R
= 100 [t2 dt] x 10 = 1000 [t3/3]1
1000 x 1/3 = 1000/3 W
- (c)
current through the surface is given by
I(t) = dQ/dt where, Q = charge flowing
Q = |I(t) dt
for 0 < t < 6 s,
Q = I (t) dt = Area under the curve I (t) – t
= Area OBC – Area BCDE
= 1/2 x 3 x 5 – [(1/2 x 1 x 1) + (1 x 1) + (1/2 x 1 x 1)]
= 5.5 C
- (a)
Energy absorbed by the device is given
E = |P(t) dt
where, p (t) = power absorbed by the device
E = |V(t) I(t) dt
= 10 (1 – e-7t) dt + |0. (e-7(t-1)) dt
= 10 (1 – e-7t) dt
= 10 (t)1 – 10 [e-7t/7]1 = 10 – 10 [-e-7t/7 + 1/7]
= 10 – 10/7 (1 – e-7t) = 8.5 j
- (c)
Applying superposition theorem,
Case 1 taken current source, circuit becomes as given in figure.
let the current flowing in R is I1
I1 = 2 x 2/2 + R
= 4/2 + R ……………….(i)
Case ii taken voltage source, circuit becomes as given in figure.
Let the current flowing is R is I2
I2 = 2/2 + R………………………(ii)
since current in both the cases flowing in the same direction, so net current
I = I1 + I2
1 = 4/2 + R + 2/2 + R (I = 1, given)
2 + R = 6
R = 6 – 2 = 4
- (a)
since, in one inductor current is leaving to dot and in other inductor current is entering to dot. so,
LQE = L1 + L2 – 2M
- (c)
by using delta to star conversion,
R1 = R12 x R13/R12 + R23 + R31 = J5 x J5/J5 + J5 – J5 = J5
R2 = R12 x R23/R12 + R23 + R31
= J5 x (-j5)/j5 + j5 – j5 = – j5
R3 = R13 x R23/R12 + R23 + R31
= J5 x – j5/j5 + j5 – j5 = – j5
- (d)
the circuit is as shown below.
100 = 65 + V2 V2 = 35 V
V3 – 30 = V2 V3 = 65 V
105 + V4 – V3 – 65 = 0 V1 = 25 V
V4 + 15 – 55 + V1 = 0 V1 = 15 V
- (b)
changing current source to voltage source,
Applying KVL,
ISRS = – I (t) RS – I (t) R – L dI(t)/dt = 0
t = 0+
I(0+) = 0 because at t = 0– there is no curent flow through inductor and inductor opposes the instantaneous flow of current in circuit and I(0–) = I(0+) = 0
dI(0+) dt = ISRS/L
- (c)
A tree is connected subgraph of a convected subgraph containing all the nodes of the graph but containing to loops.
- (d)
In pole-zero plot there are two transmission zeros are located on the jo-axis, at the complex conjugate location, then the magnitude response exhibits a zero transmission at 0 – 0.
- (b)
other three circuits can be drawn on plane without crossing as shown below.
- (a)
the circuits 1 and 2 are redrawn as below. 3 and 4 can not be redrawn on a plane without crossing other branch.
- (c)
following figure shows that afhg is a tree of given graph.
- (b)
in this graph there are 4 nodes and 6 branches.
twig = n- 1 = 4 – 1 = 3
link = b – n + 1 = 6 – 4 + 1 = 3
- (c)
using nodal analysis,
VA – 10/4 + VA/2 = 4
VA = 8.67 V
- (d)
V2/20 + V2 + 10/30 = 0.5
V2 = 2 V
- (d)
-V1/60 + -V1/60 + 6 = 9
V1 = – 90 V
- (b)
V1 = 4VS/6R + VS/3R
1/6R + 1/3R + 1/6R
V1 = 1.5 VS
- (b)
VA = 2 (3 + 1) + 3(1)
VA = 11 V
- (d)
For maximum power transfer, load impedance ZL should be complex conjugate of the series impedance ZS.
ZL = ZS
= R3 – JXS
The maximum power transfer will be
P = V2/4RS
- (c)
Impedance of capacitor XC = 1/0C
where, o = 0
XC = 0
Capacitor behaves as open-circuit.
VO = 0
XC = 0
Capacitor behaves as short-circuit.
VO = 0
Hence, RC is circuit shown is band-pass filter.
- (c)
Transfer function of circuit
Y(s)/U(s) = 1/sC/sL + R + 1/sC = 1/S2LC + sRC + 1
Dividing numerator and denominator by LC
1/LC/S2 + R/L s + 1/LC
Characteristic equation
s2 + R/L s + 1/LC = 0
Comparing with general expression,
s2 + 2E0ns + 02n = 0
on = 1/LC and 2E0N = R/L
E = 1/2 x R/L x LC
E = R/2 C/L
For response Y(t) has no oscillation;
E > 1
R > 2 L/C
- (b)
Two-port output expressed as input parameters in ABCD- parameters
[V1/V1] = [A B C D][V2/I2]
V1 = AV2 + BI2
I1 = CV2 + DI2 ………….(1)
We know that for ideal transformer,
V1/V2 = I2/I1 = N/1
V1 = nV2
I1 = I2/N
We can write V1 = nV2 + 0I2
Comparing rith eq. (i)
[A B C D] = [n 0 0 1/n]
hence, X = 1/n
- (b)
For series resonant circuit
resonant preparing fr = 1/2 LC
= 1/2 1/400 x 10-6 x 1 = 104
- (c)
For maximum power transfer to load resistor RL, RL must be equal to 100
maximum power V2/4RL = 102/4 x 100 = 0.25 W
- (d)
In order to calcuate I in the given figure draw the equivalent model.
1/VEQ = 1/15/8 = 8/15
- (a)
the constant
= REQC
REQ = 4R||4R = 2R
C = 2C
= 2R . 2C = 4RC
- (b)
we know
energy = u . N2 A/I
4 x 10-7 x (1000)2 x (3/2 x 10-2)2/30 x 10-2
= 0.15 j
- (C)
VTH = (2) (3) (1)/3 + 3 = 1 V
RTH = 1||5 = 5/6
- (b)
after killing all source equivalent resistance is R. open circuit voltage = V1.
- (a)
the circuit is as shown below.
RN = 2||4 + 2 = 10/3
V1 = 15/2/1/2 + 1/2 + 1/4 = 6 V
ISC = IN = V1/2 = 3 A
- (b)
VTH = (6) (6)/3 + 3 = 4 V
RTH = (3||6) + 2 = 4
- (d)
the short-circuit current across the terminal is
ISC = 6 x 4/4 + 2 = 4 A = IN, RN = 6||3 = 2
- (b)
For the calculation of RTH. if we kill the sources the 20 resistance is inactive because 5 A source will be open-circuit.
RTH = 30 + 25 = 55
VTH = 5 + 5 x 30 = 155 V
- (b)
A tree is connected subgraph of a connected graph containing all the nodes of the graph but containing no loop.
option (b) has loop. so, it is not a tree.
hence, (b) cannot be tree of the given graph.
- (d)
A current entering the dotted terminal of one coil produce an open-circuit voltage with positive voltage refernce at the dotted terminal of the second coil.
LEQ = dI/dt = L1 dI/dt + L2 dI/dt – M dI/dt – M dI/dt
LEQ = L1 + L2 – 2M
- (a)
V (t) = sin 2t
in phasor form,
V = 1 <00
Admittance of circuit
Y = YR + YL + YC
= 1/R + 1/f0L + J0C
= 3 + 1/jo3 x 1/4 + j x 2 x 3
= 3 – j2 + j6
= 3 + 4 j
in phasor form,
= 32 + 42 < tan-14/3
Y = 5<53.10
I = V.Y = 1<800 x 5<53.10
= 5<53.10
I = 5 sin (2t + 53.10)
- (c)
Applying KVL in loop,
V1(t) = L dI(t)/dt + I(t) R
u(t) = dI (t)/dt + 2I(t)
solution of equation
I(t) = 1/2 (1 – e-2t) u(t)
I(t) = 1/2 = 0.5
At t = 1/2.
I(t) = 1/2 (1 – e-1) = 0.31