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 V
_{O}(s)/V_{1}(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 S
_{O}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) L_{1} + L_{2} – 2 M

(b) L_{1} + L_{2} + 2 M

(c) L_{1} + L_{2} – M

(d) L_{1} + L_{2}

- 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 V
_{1}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) R_{S}I_{S}/L

(c) (R + R_{S})I_{S}/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 V
_{A}is

(a) 4.33 V

(b) 4.09 V

(c) 8.67 V

(d) 8.18 V

- In the following circuit, the voltage V
_{2}is

(a) 0.5 V

(b) 1.0 V

(c) 1.5 V

(d) 2.0 V

- In the following circuit, the voltage V
_{1}is

(a) 120 V

(b) -120 V

(c) 90 V

(d) -90 V

- In the following circuit, the voltage V
_{1}is

(a) 0.4 V_{S}

(b) 1.5 V_{S}

(c) 0.67 V_{S}

(d) 2.5 V_{S}

- In the following circuit the voltage V
_{A}is

(a) -11 V

(b) 11 V

(c) 3 V

(d) -3 V

- An independent voltage source in series with an impedance Z
_{S}= R_{S}= JX_{S}delivers a maximum average power to a load impedance Z_{L}when

(a) Z_{L} = R_{S} + JX_{S}

(b) Z_{L} = R_{S}

(c) Z_{L} = JX_{S}

(d) Z_{L} = R_{S} = JX_{S}

- 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) n^{2}

(d) 1/n^{2}

- In a series RLC circuit R = 2 K, L = 1 H and C = 1/400 uf. the resonant frequency is

(a) 2 x 10^{4 }hz

(b) 1 x 10^{4} hz

(c) 10^{4} hz

(d) 2 x 10^{4} hz

- The maximum power that can be transferred to the load resistor R
_{L}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 V
_{TH}and R_{TH}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 I
_{N}and R_{N}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 V
_{TH}and R_{TH}are

(a) 2 V, 4

(b) 4 V, 4

(c) 4 V, 5

(d) 2 V, 5

- In the following circuit, the values of I
_{N}and R_{N}are

(a) 4 A, 3

(b) 2 A, 6

(c) 2 A, 9

(d) 4 A, 2

- For the following circuit, the values of V
_{TH}and R_{TH}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) L_{1} + L_{2} + M

(b) L_{1} + L_{2} – M

(c) L_{1} + L_{2} + 2M

(d) L_{1} + L_{2} – 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 + 53.1^{0})

(b) 5 sin (2t + – 53.1^{0})

(c) 25 sin (2t + 53.1^{0})

(d) 25 sin (2t – 53.1^{0})

- 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

[I_{1}/I_{2}] = [Y_{11} Y_{12}/Y_{21} Y_{22}] [V_{1}/V_{2}]

I_{1} = Y_{11} V_{1} + Y_{12} V_{2}

I_{2} = Y_{22} Y_{1} + Y_{22} V_{2}

Y_{11} = I_{1}/V_{1}|_{V2 = 0}

Y_{12} = I_{1}/V_{2}|_{V1 = 0}

Y_{21} = I_{2}/V_{1}|_{V2 = 0}

Y_{22} = I_{2}/V_{2}|_{V1 = 0}

**To calcuate Y _{11} and Y_{21}**

Terminals 2 – 2′ make shorten V_{2} = 0

Now, circuit is

Y_{11 }= I_{1}/V_{1}|_{V2 = 0} = 1/0.5110 x 5 = 1/0.25 = 4

V_{1} = – 0.5 I_{2}

I_{2}/V_{1} = – 1/0.5 I_{2}

I_{2}/V_{1} = – 1/0.5 = – 2

Y_{21} = I_{2}/V_{1} = – 2

**To calcuate Y _{12} and Y_{22}**

Terminals 1 – 1′ make shorten as V_{1} = 0.

Now, circuit is

V_{2} = 0.5I_{1}

Y_{12} = I_{1}/V_{2}|_{V1 = 0} = 1/-0.5 = – 2

V_{2} = (0.5110.5)I_{2} = 0.25I_{2}

Y_{22} = I_{2}/V_{2}|_{V1 = 0} = 1/0.25 = 4

Hence, [Y_{11} Y_{12}/Y_{21} Y_{22}] = [4 -2/-2 4]

**Alternate Method**

For general network

The admittance parameters

Y_{11} = Y_{A} + Y_{C}

Y_{12} = – Y_{C}

Y_{21} = – Y_{C}

Y_{22} = Y_{C} + Y_{B}

Convert impedance to admittance in circuit

Y_{11} = 2 + 2 = 4

Y_{12} = – 2

Y_{21} = – 2

Y_{22} = 2 + 2 = 4

[Y_{11} Y_{12}/Y_{21} Y_{22}] = [4 -2 /-2 4]

- (d)

parallel resonant circuit

Q = 0_{0}/s_{0}

so = o_{0}/Q

- O
_{O}RC = R/0_{O}L = R/R C/L

s_{o} = B_{O} = 1/LC x 1L/RL = 1/RC

S_{o} = B_{O} = 1/LC x 1/L/RC = 1/RC

O_{0} = 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/C_{EQ} = 1/C + 1/C + 1/5C/3

= 1/C + 1/C + 3/5C = 1/C [5 + 5 + 3/5]

C_{EQ} = 5C/13

- (a)

Suppose value of resistance be R each.

Applying KVL by selecting any path between A and B,

V_{AB} = 1/3. R + 1/6 R + 1/3 R

V_{AB} = IR [5/6]

V_{AB}/I = 5R/6

R_{AB} = 5R/6

Where, R = 1 for this problem

so, R_{AB} = 5/6

- (a)

Apply same concept as done in earlier problem.

- (a)

R_{AB }= 1 + (2||R_{AB})

R_{AB} = 1 + 2.R_{AB}/2 + R_{AB}

R_{AB} = 2 + R_{AB} + 2R_{AB}/2 + R_{AB}

2R_{AB} + (R_{AB})^{2} = 2 + R^{AB} + 2R_{AB}

(R_{AB})^{2} – R_{AB} – 2 = 0

R^{2}_{AB} – 2R_{AB} + R_{AB} – 2 = 0

R_{AB} (R_{AB} – 2) + 1 (R_{AB} – 2) = 0

(R_{AB} + 1) (R_{AB} – 2) = 0

R_{AB} = – 1,2

R_{AB} = – 1not possible

R_{AB} = 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 I_{1} in figure)

I + I_{1} = 12 A ………….(i)

current source must be have the value less than 12 A to satisfies eq . (1).

- (c)

Impedance of capacitor X_{C} = 1/sC

V_{O} = I_{L}R_{I ………………….}(i)

From current division rule,

I_{L} = I x X_{C}/R_{L} + X_{C}

V_{O} = I x CR_{L}/R_{L} + X_{C} ………..(ii)

Equivalent impedance of circuit

Z_{EQ} = R + (XC||R_{L})

= R + X_{C}R_{L}/X_{C}R_{L}

We can write

I = V_{I}/Z_{EQ}

Putting value in eq. (ii)

V_{O} (S) = V_{I} (s)/X_{C}R_{L} x X_{C}R_{L}/X_{C} + R_{L}

V_{O}(s)/V_{I} (s) = X_{C}R_{L}/RX_{C} + RR_{L} + X_{C}R_{L}

Dividing by X_{C}R_{L},

V_{O}(s)/V_{I} (s) = 1/1 + R/X_{C} + R/R_{L} = 1/1 + RsC + R/R_{L}

R = R_{L}

V_{O}/V_{I} (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 |I^{2} dt]. R

= 100 [t^{2} dt] x 10 = 1000 [t^{3}/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 I_{1}

I_{1} = 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 I_{2}

I_{2} = 2/2 + R………………………(ii)

since current in both the cases flowing in the same direction, so net current

I = I_{1} + I_{2}

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,

L_{QE} = L_{1} + L_{2} – 2M

- (c)

by using delta to star conversion,

R_{1} = R_{12} x R_{13}/R_{12 }+ R_{23} + R_{31} = J5 x J5/J5 + J5 – J5 = J5

R_{2 } = R_{12} x R_{23}/R_{12} + R_{23} + R_{31}

= J5 x (-j5)/j5 + j5 – j5 = – j5

R_{3} = R_{13} x R_{23}/R_{12} + R_{23} + R_{31}

= J5 x – j5/j5 + j5 – j5 = – j5

- (d)

the circuit is as shown below.

100 = 65 + V_{2} V_{2} = 35 V

V_{3} – 30 = V_{2} V_{3} = 65 V

105 + V_{4} – V_{3} – 65 = 0 V_{1} = 25 V

V_{4} + 15 – 55 + V_{1} = 0 V_{1} = 15 V

- (b)

changing current source to voltage source,

Applying KVL,

I_{S}R_{S }= – I (t) R_{S} – 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 = I_{S}R_{S}/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,

V_{A} – 10/4 + V_{A}/2 = 4

V_{A} = 8.67 V

- (d)

V_{2}/20 + V_{2} + 10/30 = 0.5

V_{2 }= 2 V

- (d)

-V_{1}/60 + -V_{1}/60 + 6 = 9

V_{1} = – 90 V

- (b)

V_{1} = 4V_{S}/6R + V_{S}/3R

1/6R + 1/3R + 1/6R

V_{1} = 1.5 V_{S}

- (b)

V_{A} = 2 (3 + 1) + 3(1)

V_{A} = 11 V

- (d)

For maximum power transfer, load impedance Z_{L} should be complex conjugate of the series impedance Z_{S}.

Z_{L} = Z_{S}

= R_{3} – JX_{S}

The maximum power transfer will be

P = V^{2}/4R_{S}

- (c)

Impedance of capacitor X_{C} = 1/0C

where, o = 0

X_{C} = 0

Capacitor behaves as open-circuit.

V_{O} = 0

X_{C } = 0

Capacitor behaves as short-circuit.

V_{O} = 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/S^{2}LC + sRC + 1

Dividing numerator and denominator by LC

1/LC/S^{2} + R/L s + 1/LC

Characteristic equation

s^{2} + R/L s + 1/LC = 0

Comparing with general expression,

s^{2} + 2E0_{n}s + 0^{2}_{n } = 0

o_{n} = 1/LC and 2E0_{N} = 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

[V_{1}/V_{1}] = [A B C D][V_{2}/I_{2}]

V_{1} = AV_{2} + BI_{2 }

I_{1} = CV_{2} + DI_{2} ………….(1)

We know that for ideal transformer,

V_{1}/V_{2} = I_{2}/I_{1} = N/1

V_{1} = nV_{2}

I_{1} = I_{2}/N

We can write V_{1} = nV_{2} + 0I_{2}

Comparing rith eq. (i)

[A B C D] = [n 0 0 1/n]

hence, X = 1/n

- (b)

For series resonant circuit

resonant preparing f_{r} = 1/2 LC

= 1/2 1/400 x 10^{-6} x 1 = 10^{4}

- (c)

For maximum power transfer to load resistor R_{L}, R_{L} must be equal to 100

maximum power V^{2/}4R_{L} = 10^{2}/4 x 100 = 0.25 W

- (d)

In order to calcuate I in the given figure draw the equivalent model.

1/V_{EQ} = 1/15_{/}8 = 8/15

- (a)

the constant

= R_{EQ}C

R_{EQ} = 4R||4R = 2R

C = 2C

= 2R . 2C = 4RC

- (b)

we know

energy = u . N^{2} A/I

4 x 10^{-7 }x (1000)^{2} x (3/2 x 10^{-2})^{2}/30 x 10^{-2}

= 0.15 j

- (C)

V_{TH} = (2) (3) (1)/3 + 3 = 1 V

R_{TH} = 1||5 = 5/6

- (b)

after killing all source equivalent resistance is R. open circuit voltage = V_{1.}

- (a)

the circuit is as shown below.

R_{N} = 2||4 + 2 = 10/3

V_{1} = 15_{/}2/1_{/}2 + 1_{/}2 + 1_{/}4 = 6 V

I_{SC} = I_{N} = V_{1}/2 = 3 A

- (b)

V_{TH} = (6) (6)/3 + 3 = 4 V

R_{TH} = (3||6) + 2 = 4

- (d)

the short-circuit current across the terminal is

I_{SC} = 6 x 4/4 + 2 = 4 A = I_{N}, R_{N} = 6||3 = 2

- (b)

For the calculation of R_{TH}. if we kill the sources the 20 resistance is inactive because 5 A source will be open-circuit.

R_{TH} = 30 + 25 = 55

V_{TH} = 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.

L_{EQ} = dI/dt = L_{1 } dI/dt + L_{2 }dI/dt – M dI/dt – M dI/dt

L_{EQ} = L_{1 }+ L_{2} – 2M

- (a)

V (t) = sin 2t

in phasor form,

V = 1 <0^{0}

Admittance of circuit

Y = Y_{R} + Y_{L} + Y_{C}

= 1/R + 1/f0L + J0C

= 3 + 1/j_{o}3 x 1/4 + j x 2 x 3

= 3 – j2 + j6

= 3 + 4 j

in phasor form,

= 3^{2} + 4^{2 }< tan^{-1}4/3

Y = 5<53.1^{0}

I = V.Y = 1<80^{0} x 5<53.1^{0}

= 5<53.1^{0}

I = 5 sin (2t + 53.1^{0})

- (c)

Applying KVL in loop,

V_{1}(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