Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

101.

The value of the Y-parameter Y22 in the circuit shown below.(a) 12/7(b) 6/7(c) 7/6(d) 7/12I have been asked this question by my college director while I was bunking the class.The query is from Terminated Two-Port Network in division Two-Port Networks of Network Theory

Answer»

The correct choice is (d) 7/12

Easiest explanation: The relation between Y22 and I2/V2 is Y22 = I2/V2. We have the relation I2/V2 = (Y22Ys+Y22Y11-Y21Y12)/(Ys+Y11). On substituting their values in the EQUATION we GET Y22 = 7/12 MHO.

102.

Find the value of I2/V2 in the circuit shown below.(a) 7/6(b) 6/7(c) 7/12(d) 12/7The question was asked in semester exam.My enquiry is from Terminated Two-Port Network in portion Two-Port Networks of Network Theory

Answer»

The CORRECT OPTION is (c) 7/12

Easiest EXPLANATION: The RELATION between I2/V2 and Y-parameters is

 I2/V2=(5/8×1+5/8×1/2-1/16)/(1+1/2)=7/12 mho.

103.

Determine the value of source admittance in the circuit shown below.(a) 1(b) 2(c) 3(d) 4I had been asked this question in semester exam.The above asked question is from Terminated Two-Port Network topic in portion Two-Port Networks of Network Theory

Answer»

Right CHOICE is (a) 1

Explanation: From the figure, the value of the admittance parallel to the CURRENT source is 1 mho and this is the value of source admittance. So Ys = 1 mho.

104.

Determine the input impedance of the network shown below.(a) 4.25(b) 3.25(c) 2.25(d) 1.25I had been asked this question in a job interview.My doubt is from Terminated Two-Port Network in chapter Two-Port Networks of Network Theory

Answer»

Correct choice is (b) 3.25

For EXPLANATION: From the figure by INSPECTION we can say that the source RESISTANCE is 1Ω. So ZIN = (V1/I1) + Source resistance. We had V1/I1 = 2.25. On substituting Zin=1+2.25=3.25Ω.

105.

Find the value of V1/I1 in the circuit shown below.(a) 1.25(b) 2.25(c) 3.25(d) 4.25This question was addressed to me in my homework.My enquiry is from Terminated Two-Port Network in division Two-Port Networks of Network Theory

Answer»

The CORRECT ANSWER is (b) 2.25

The BEST I can EXPLAIN: We have the relation V1/I1=Z11– Z12Z21/(ZL+Z21) and ZL is the LOAD impedance and is equal to 2Ω. On solving V1/I1=2.5-1/(2+2)=2.25Ω.

106.

Determine the Z-parameter Z22 in the circuit shown below.(a) 1(b) 3(c) 2(d) 4I have been asked this question by my college director while I was bunking the class.My question is from Terminated Two-Port Network in chapter Two-Port Networks of Network Theory

Answer»

The correct choice is (C) 2

Easy explanation: The Z-parameter Z21 is V2/I2 |I1=0. This parameter is OBTAINED by open CIRCUITING port 1. So we get V2 = ((2+2)||4)I2 => V2 = 2(I2) => V2/I2 = 2. On SUBSTITUTING Z21 = 2Ω.

107.

Determine the Z-parameter Z21 in the circuit shown below.(a) 4(b) 3(c) 2(d) 1I had been asked this question in class test.This is a very interesting question from Terminated Two-Port Network in portion Two-Port Networks of Network Theory

Answer»

Correct ANSWER is (d) 1

The best explanation: The Z-parameter Z21 is V2/I1 |I2=0. On open CIRCUITING PORT 2, we GET V2 = (1.5)I1 => V2/I1 = 1.5. On substituting we get Z21 = 1.5Ω.

108.

Determine the Z-parameter Z12 in the circuit shown below.(a) 1(b) 2(c) 3(d) 4The question was asked during an internship interview.My question is based upon Terminated Two-Port Network topic in division Two-Port Networks of Network Theory

Answer»

Correct answer is (a) 1

Easy explanation: The Z-parameter Z12 is V2/I1 |I2=0. On OPEN CIRCUITING port 2 we obtain the equation, V1 = (1.5) I2 => V1/I1 = 1.5. On substituting we get Z12 = 1.5Ω.

109.

From the circuits shown below, find the combined Z-parameter Z21.(a) 7(b) 6(c) 5(d) 4The question was posed to me in exam.My question comes from Inter Connection of Two-Port Networks topic in chapter Two-Port Networks of Network Theory

Answer»

The correct CHOICE is (a) 7

The BEST EXPLANATION: The Z-parameter Z21 is Z21 = Z21x + Z21y and we have Z21x = 2, Z21y = 5. On SUBSTITUTING we get Z21 = 2 + 5 = 7Ω.

110.

From the circuits shown below, find the combined Z-parameter Z22.(a) 38(b) 28(c) 18(d) 8This question was addressed to me in class test.I would like to ask this question from Inter Connection of Two-Port Networks in chapter Two-Port Networks of Network Theory

Answer»

Right choice is (b) 28

To EXPLAIN I WOULD say: The Z-parameter Z22 is Z22 = Z22x + Z22y and we have Z22x = 3, Z22y = 25. On substituting we GET Z22 = 3 +25 = 28Ω.

111.

Calculate the Z –parameter Z11 in the circuit shown below.(a) 1.5(b) 2.5(c) 3.5(d) 4.5I have been asked this question during an interview.My question comes from Terminated Two-Port Network topic in portion Two-Port Networks of Network Theory

Answer»

Right answer is (b) 2.5

Explanation: The Z–parameter Z11 is V1/I1, PORT 2 is open circuited. V1 = (1+1.5)I1 => V1/I1 = 2.5 and on SUBSTITUTING, we GET Z11 = 2.5Ω.

112.

From the circuits shown below, find the combined Z-parameter Z11.(a) 8(b) 18(c) 28(d) 38I got this question in an interview for job.This intriguing question originated from Inter Connection of Two-Port Networks topic in section Two-Port Networks of Network Theory

Answer» CORRECT option is (B) 18

The explanation: The Z-parameter Z11 is Z11 = Z11x + Z11y and Z11x = 3, Z11y = 15. On substituting we GET Z11 = 3 +15 = 18Ω.
113.

From the circuits shown below, find the combined Z-parameter Z12.(a) 4(b) 5(c) 6(d) 7I have been asked this question in final exam.The above asked question is from Inter Connection of Two-Port Networks topic in section Two-Port Networks of Network Theory

Answer»

The CORRECT answer is (d) 7

The EXPLANATION is: The Z-parameter Z12 is Z12 = Z12x + Z12y and we have Z12x = 2, Z12y. On SUBSTITUTING we GET Z12 = 2 + 5 = 7Ω.

114.

In the circuit shown below, find the Z-parameter Z11.(a) 10(b) 15(c) 20(d) 25This question was addressed to me during a job interview.My question comes from Inter Connection of Two-Port Networks in chapter Two-Port Networks of Network Theory

Answer»

The correct option is (b) 15

The explanation: The Z –parameter Z11 is V1/I1, PORT 2 is OPEN circuited. V1 = (10 + 5)I1 => V1/I1 = 15 and on SUBSTITUTING, we GET Z11 = 2.5Ω.

115.

In the circuit shown below, find the Z-parameter Z12.(a) 15(b) 10(c) 5(d) 1I have been asked this question in an online interview.This is a very interesting question from Inter Connection of Two-Port Networks in section Two-Port Networks of Network Theory

Answer»

The CORRECT answer is (c) 5

To explain I WOULD say: The Z-parameter Z12 is V2/I1 |I2=0. On open circuiting port 2 we obtain the EQUATION, V1 = (5) I2 => V1/I1 = 5. On substituting we get Z12 = 5Ω.

116.

In the circuit shown below, find the Z-parameter Z22.(a) 3(b) 2(c) 4(d) 1I got this question at a job interview.Asked question is from Inter Connection of Two-Port Networks in chapter Two-Port Networks of Network Theory

Answer»

Right OPTION is (a) 3

Easiest explanation: The Z-parameter Z21 is V2/I2 |I1=0. This parameter is obtained by open circuiting port 1. So we GET V2 = (2 + 1)I2 => V2 = 3(I2) => V2/I2 = 3. On substituting Z21 = 3Ω.

117.

In the circuit shown below, find the Z-parameter Z21.(a) 2(b) 4(c) 1(d) 3I got this question in an interview for job.My question is from Inter Connection of Two-Port Networks in section Two-Port Networks of Network Theory

Answer»

The correct ANSWER is (a) 2

For EXPLANATION: The Z-parameter Z21 is V2/I1 |I2=0. On open circuiting PORT 2, we get V2 = (2)I1 => V2/I1 = 2. On substituting we get Z21 = 2Ω.

118.

In the circuit shown below, find the Z-parameter Z11.(a) 1(b) 2(c) 3(d) 4This question was posed to me in examination.Enquiry is from Inter Connection of Two-Port Networks in portion Two-Port Networks of Network Theory

Answer»

Correct answer is (c) 3

Best EXPLANATION: The Z –parameter Z11 is V1/I1, port 2 is open circuited. V1 = (1+2)I1 => V1/I1 = 3 and on SUBSTITUTING, we get Z11 = 3Ω.

119.

In the circuit shown below, find the Z-parameter Z12.(a) 4(b) 3(c) 2(d) 1This question was posed to me in class test.I want to ask this question from Inter Connection of Two-Port Networks topic in chapter Two-Port Networks of Network Theory

Answer»

Right choice is (c) 2

Easy explanation: The Z-parameter Z12 is V2/I1 |I2=0. On OPEN CIRCUITING port 2 we obtain the equation, V1 = (2) I2 => V1/I1 = 2. On substituting we GET Z12 = 2Ω.

120.

The relation between Z22 and Y parameters is?(a) Z22 = (-Y11)/∆y(b) Z22 = Y21/∆y(c) Z22 = (-Y21)/∆y(d) Z22 = Y11/∆yI have been asked this question during an internship interview.I'd like to ask this question from Inner Relationships of Different Parameters in section Two-Port Networks of Network Theory

Answer» RIGHT option is (d) Z22 = Y11/∆y

Easy explanation: V2=(-Y21/∆y)I1+(Y11/∆y)I2. The RELATION between Z22 and Y parameters is Z22 = Y11/∆y.
121.

The relation between Z21 and Y parameters is?(a) Z21 = Y21/∆y(b) Z21 = Y12/∆y(c) Z21 = (-Y21)/∆y(d) Z21 = (-Y12)/∆yThe question was asked in quiz.Origin of the question is Inner Relationships of Different Parameters topic in division Two-Port Networks of Network Theory

Answer»

Right choice is (c) Z21 = (-Y21)/∆y

For explanation: V2=(-Y21/∆y)I1+(Y11/∆y)I2. The RELATION between Z21 and Y PARAMETERS is Z21 = (-Y21)/∆y.

122.

The relation between Z12 and Y parameters is?(a) Z12 = Y12/∆y(b) Z12 = (-Y12)/∆y(c) Z12 = (-Y22)/∆y(d) Z12 = Y22/∆yI had been asked this question by my school principal while I was bunking the class.My question is from Inner Relationships of Different Parameters in division Two-Port Networks of Network Theory

Answer»

The correct CHOICE is (B) Z12 = (-Y12)/∆y

The EXPLANATION: V1=(Y22/∆y)I1-(Y12/∆y)I2. The relation between Z12 and Y parameters is Z12 = (-Y12)/∆y.

123.

The relation between Z11 and Y parameters is?(a) Z11 = Y22/∆y(b) Z11 = -Y22/∆y(c) Z11 = Y12/∆y(d) Z11 = (-Y12)/∆yThe question was asked in an online quiz.I'm obligated to ask this question of Inner Relationships of Different Parameters topic in division Two-Port Networks of Network Theory

Answer»

Right option is (a) Z11 = Y22/∆y

Easy EXPLANATION: V1=(Y22/∆y)I1-(Y12/∆y)I2. The relation between Z11 and Y parameters is Z11 = Y22/∆y.

124.

For the given information Z11 = 3, Z12 = 1, Z21 = 2, Z22 = 1. What is the product of ∆y and ∆z is?(a) 3(b) 2(c) 1(d) 0I have been asked this question in exam.This intriguing question comes from Inner Relationships of Different Parameters in chapter Two-Port Networks of Network Theory

Answer»

The CORRECT choice is (c) 1

To ELABORATE: ∆y is the determinant of y parameters and ∆Z is the determinant of z parameters. And we obtained ∆y = 1 and ∆z = 1. So their PRODUCT = (1) (1) = 1.

125.

For the given information Z11 = 3, Z12 = 1, Z21 = 2, Z22 = 1. What is the value ∆y?(a) 0(b) 1(c) 2(d) 3This question was addressed to me by my college professor while I was bunking the class.This question is from Inner Relationships of Different Parameters in portion Two-Port Networks of Network Theory

Answer»

The CORRECT answer is (b) 1

For explanation I would say: ∆y is the DETERMINANT of y parameters. The value ∆y is (Y11)(Y22)-(Y12)(Y21). On substituting the values we GET ∆y = (1)(3)-(-2)(-1)=1.

126.

For the given information Z11 = 3, Z12 = 1, Z21 = 2, Z22 = 1. Find the value of Y22.(a) 1(b) 2(c) 3(d) 4This question was addressed to me in an interview.The origin of the question is Inner Relationships of Different Parameters in division Two-Port Networks of Network Theory

Answer» RIGHT OPTION is (c) 3

Best explanation: The RELATION between Y22 and Z11 is Y22= Z11/∆z and ∆z=3-2=1 and given Z11 = 3. On substituting we get Y22 = 3/1 = 3.
127.

For the given information Z11 = 3, Z12 = 1, Z21 = 2, Z22 = 1. Find the value of Y21.(a) 2(b) -2(c) 1(d) -1The question was posed to me in examination.I'd like to ask this question from Inner Relationships of Different Parameters topic in chapter Two-Port Networks of Network Theory

Answer» CORRECT option is (b) -2

For explanation I would say: We have the relation Y21=-Z21/∆z. ∆z=3-2=1 and GIVEN Z21 = 2. On substituting we get Y21=-2/1=-2.
128.

For the given information Z11 = 3, Z12 = 1, Z21 = 2, Z22 = 1. Find the value of Y12.(a) -2(b) 2(c) -1(d) 1I have been asked this question in homework.I'm obligated to ask this question of Inner Relationships of Different Parameters in division Two-Port Networks of Network Theory

Answer»

Correct OPTION is (C) -1

To ELABORATE: Y12 = -Z12/∆z and ∆z=3-2=1 and Z12 = 1. So on substituting we get Y12=-1/1=-1.

129.

For the given information Z11 = 3, Z12 = 1, Z21 = 2, Z22 = 1. Find the value of Y11.(a) 1(b) -1(c) 2(d) -2This question was addressed to me in an interview for job.The question is from Inner Relationships of Different Parameters topic in section Two-Port Networks of Network Theory

Answer» CORRECT answer is (a) 1

For EXPLANATION: Y11 = Z22/∆z and ∆z=3-2=1 and Z22 = 1. So on substituting we GET Y11 = 1/1 = 1.
130.

For the circuit given below, the value of g11 and g21 are _________________(a) g11 = 0.0667 – j0.0333 Ω, g21 =0.8 + j0.4 Ω(b) g11 = -0.0667 – j0.0333 Ω, g21 = -0.8 – j0.4 Ω(c) g11 = 0.0667 + j0.0333 Ω, g21 = 0.8 + j0.4 Ω(d) g11 = -0.0667 + j0.0333 Ω, g21 = 0.8 – j0.4 ΩI had been asked this question during an online interview.This intriguing question comes from Advanced Problems Involving Parameters topic in portion Two-Port Networks of Network Theory

Answer» RIGHT option is (C) g11 = 0.0667 + j0.0333 Ω, G21 = 0.8 + j0.4 Ω

For EXPLANATION I would SAY: V1 = (12-j6) I1

Or, g11 = \(\frac{I_1}{V_1} = \frac{1}{12-j6}\) = 0.0667 + j0.0333 Ω

g21 = \(\frac{V_2}{V_1}= \frac{12I_1}{(12-j6) I_1}\)

= \(\frac{2}{2-j}\) = 0.8 + j0.4 Ω.
131.

For the circuit given below, the value of g12 and g22 are ________________(a) g12 = 0.8 + j0.4 Ω, g22 = 2.4 + j5.2 Ω(b) g12 = -0.8 + j0.4 Ω, g22 = -2.4 – j5.2 Ω(c) g12 = 0.8 – j0.4 Ω, g22 = 2.4 – j5.2 Ω(d) g12 = -0.8 – j0.4 Ω, g22 = 2.4 + j5.2 ΩThis question was posed to me by my school teacher while I was bunking the class.This question is from Advanced Problems Involving Parameters topic in division Two-Port Networks of Network Theory

Answer»

The correct option is (d) g12 = -0.8 – j0.4 Ω, g22 = 2.4 + j5.2 Ω

Explanation: I1 = \(\frac{-12}{12-J6}\)I2

Or, g12 = \(\frac{I_1}{I_2}\) = -g21 = -0.8 – j0.4 Ω

V2 = (J10 + 12 || -j6) I2

g22 = \(\frac{V_2}{I_2}\) = 2.4 + j5.2 Ω.

132.

For the circuit given below, the value of the g12 and g22 are _______________(a) g12 = –\(\frac{R_2}{R_1+R_2}\), g22 = R3 + \(\frac{R_1 R_2}{R_1+R_2}\)(b) g12 = \(\frac{R_2}{R_1+R_2}\), g22 = -R3 + \(\frac{R_1 R_2}{R_1+R_2}\)(c) g12 = –\(\frac{R_2}{R_1+R_2}\), g22 = R3 – \(\frac{R_1 R_2}{R_1+R_2}\)(d) g12 = \(\frac{R_2}{R_1+R_2}\), g22 = -R3 – \(\frac{R_1 R_2}{R_1+R_2}\)The question was posed to me during an interview.Asked question is from Advanced Problems Involving Parameters in division Two-Port Networks of Network Theory

Answer» RIGHT ANSWER is (a) g12 = –\(\FRAC{R_2}{R_1+R_2}\), g22 = R3 + \(\frac{R_1 R_2}{R_1+R_2}\)

The EXPLANATION is: I1 = –\(\frac{R_2}{R_1+R_2}\)I2

Or, g12 = \(\frac{I_1}{I_2} = -\frac{R_2}{R_1+R_2}\)

Also, I2 (R3 + R1 //R2)

= I2 \((R_3 + \frac{R_1 R_2}{R_1+R_2})\)

∴ g22 = \(\frac{V_2}{I_2} = R_3 + \frac{R_1 R_2}{R_1+R_2}\).
133.

For a 2-port network, the value of the h parameter is as h=[600, 0.04; 30, 2×10^-3]. Given that, ZS = 2 kΩ and ZL = 400 Ω. The value of the parameter Zout is ______________(a) 650 Ω(b) 500 Ω(c) 250 Ω(d) 600 ΩThe question was posed to me in an internship interview.My doubt stems from Advanced Problems Involving Parameters topic in portion Two-Port Networks of Network Theory

Answer»

The correct option is (a) 650 Ω

The best explanation: Zout = \(\frac{R_s+h_{IE}}{(R_s+h_{ie}) h_{oe}-h_{re} h_{FE}}\)

= \(\frac{R_s+h_{11}}{(R_s+h_{11}) h_{22}-h_{21} h_{12}}\)

= \(\frac{2000+600}{2600×2×10^{-3}-30×0.04}\) = 650 Ω.

134.

For the circuit given below, the value of the g11 and g21 are _________________(a) g11 = –\(\frac{1}{R_1+R_2}\), g21 = \(\frac{R_2}{R_1+R_2}\)(b) g11 = \(\frac{1}{R_1-R_2}\), g21 = –\(\frac{R_2}{R_1+R_2}\)(c) g11 = \(\frac{1}{R_1+R_2}\), g21 = \(\frac{R_2}{R_1+R_2}\)(d) g11 = \(\frac{1}{R_1-R_2}\), g21 = \(\frac{R_2}{R_1-R_2}\)I got this question in examination.Origin of the question is Advanced Problems Involving Parameters in chapter Two-Port Networks of Network Theory

Answer»

The CORRECT option is (C) G11 = \(\frac{1}{R_1+R_2}\), G21 = \(\frac{R_2}{R_1+R_2}\)

To explain I would SAY: I1 = \(\frac{V_1}{R_1+R_2}\)

Or, g11 = \(\frac{I_1}{V_1}= \frac{1}{R_1+R_2}\)

By voltage division, V2 = \(\frac{R_2}{R_1+R_2}\)V1

Or, g21 = \(\frac{V_2}{V_1} = \frac{R_2}{R_1+R_2}\).

135.

For a 2-port network, the value of the h parameter is as h=[600, 0.04; 30, 2×10^-3]. Given that, ZS = 2 kΩ and ZL = 400 Ω. The value of the parameter Zin is ______________(a) 250 Ω(b) 333.33 Ω(c) 650 Ω(d) 600 ΩThe question was asked during a job interview.My enquiry is from Advanced Problems Involving Parameters in portion Two-Port Networks of Network Theory

Answer»

The CORRECT choice is (b) 333.33 Ω

To explain I WOULD SAY: Zin = hie – \(\frac{h_{re} h_{fe} R_L}{1 + h_{oe} R_L}\)

= h11 – \(\frac{h_{12} h_{21} R_L}{1+h_{22} R_L}\)

= 600 – \(\frac{0.04×30×400}{1+2×10^{-3}×400}\) = 333.33 Ω.

136.

If for a circuit the value of the h parameter is given as h = [8, 2/3; -2/3, 4/9]. Then the value of the voltage source V is _________________(a) 2.38 V(b) 1.19 V(c) 1.6 V(d) 3.2 VI had been asked this question in final exam.The above asked question is from Advanced Problems Involving Parameters in section Two-Port Networks of Network Theory

Answer»

Right choice is (b) 1.19 V

The best explanation: 8I1 + \(\frac{2}{3V_2}\) = 10

V2 = \(\frac{2}{3}\)I1 (5||\(\frac{9}{4}\))

= \(\frac{2}{3}\)I1 \((\frac{45}{29})= \frac{30}{29}I_1\)

I1 = \(\frac{29}{30}\)V2

(8)(\(\frac{29}{30}\)) V2 + \(\frac{2}{3}\)V2 = 10

V2 = \(\frac{300}{252}\) = 1.19 V.

137.

For the circuit given below, the values of the h parameter is given as follows h = [16, 3; -2, 0.01]. The value of the ratio \(\frac{I_2}{I_1}\)is ______________(a) 0.3299(b) 0.8942(c) -1.6(d) 0.2941The question was asked during an internship interview.My query is from Advanced Problems Involving Parameters in section Two-Port Networks of Network Theory

Answer»

The correct answer is (c) -1.6

The explanation: Replacing the given 2-port circuit by an equivalent circuit and applying nodal ANALYSIS, we get,

V2 = (20) (2I1) = 40 I1

Or, -10 + 20I1 + 3V2 = 0

Or, 10 = 20I1 + (3) (40I1) = 140I1

∴ I1 = \(\frac{1}{14}\) and V2 = \(\frac{40}{14}\)

So, V1 = 16I1 + 3V2 = \(\frac{136}{14}\)

And I2 = (\(\frac{100}{125}\)) (2I1) = \(\frac{-8}{70}\)

∴ \(\frac{I_2}{I_1}\) = -1.6.

138.

For the circuit given below, the values of the h parameter is given as follows h = [16, 3; -2, 0.01]. The value of the ratio \(\frac{V_2}{V_1}\)is ______________(a) 0.3299(b) 0.8942(c) 1.6(d) 0.2941This question was posed to me during an internship interview.I'd like to ask this question from Advanced Problems Involving Parameters topic in division Two-Port Networks of Network Theory

Answer»

Right option is (d) 0.2941

To elaborate: Replacing the given 2-port circuit by an equivalent circuit and applying nodal ANALYSIS, we get,

V2 = (20) (2I1) = 40 I1

Or, -10 + 20I1 + 3V2 = 0

Or, 10 = 20I1 + (3) (40I1) = 140I1

∴ I1 = \(\frac{1}{14}\) and V2 = \(\frac{40}{14}\)

So, V1 = 16I1 + 3V2 = \(\frac{136}{14}\)

And I2 = (\(\frac{100}{125}\)) (2I1) = \(\frac{-8}{70}\)

∴ \(\frac{V_2}{V_1}= \frac{40}{136}\) = 0.2941.

139.

For the circuit given below, the value of the Transmission parameters B and D are ________________(a) B = Y, D = 1(b) B = 1, D = 0(c) B = 0, D = 1(d) B = 0, D = YI got this question by my school principal while I was bunking the class.I'd like to ask this question from Advanced Problems Involving Parameters topic in section Two-Port Networks of Network Theory

Answer»

The correct option is (c) B = 0, D = 1

For explanation I would SAY: V1 = V2 = 0

And I2 = -I1

∴ B = \(\frac{V_1}{I_2}\) = 0

∴ D = \(\frac{-I_1}{I_2}\) = 1.

140.

For the circuit given below, the value of the Transmission parameters A and C are _______________(a) A = 1, C = 0(b) A = 0, C = 1(c) A = Y, C = 1(d) A = 1, C = YThis question was addressed to me in examination.I'm obligated to ask this question of Advanced Problems Involving Parameters topic in division Two-Port Networks of Network Theory

Answer» RIGHT ANSWER is (d) A = 1, C = Y

Explanation: V1 = V2

∴ A = \(\frac{V_1}{V_2}\) = 1

And V1 = ZI1

∴ C = \(\frac{I_1}{V_2} = \frac{1}{Z}\) = Y.
141.

For the circuit given below, the value of the Transmission parameters B and D are _________________(a) B = Z, D = 1(b) B = 1, D = Z(c) B = Z, D = Z(d) B = 1, D = 1This question was addressed to me in an international level competition.The query is from Advanced Problems Involving Parameters topic in division Two-Port Networks of Network Theory

Answer»

Correct OPTION is (a) B = Z, D = 1

The BEST I can explain: V1 = ZI1

And I2 = -I1

B = \(\frac{V_1}{I_2}\)

= \(\frac{-ZI_1}{-I_1}\)= Z

D = \(\frac{-I_1}{I_2}\)= 1.

142.

For the circuit given below, the value of the Transmission parameters A and C are _________________(a) A = 0, C = 1(b) A = 1, C = 0(c) A = Z, C = 1(d) A = 1, C = ZThe question was asked in a national level competition.The query is from Advanced Problems Involving Parameters in section Two-Port Networks of Network Theory

Answer»

Right ANSWER is (B) A = 1, C = 0

The explanation: V1 = V2

Or, A = \(\frac{V_1}{I_2}\)= 1

I1 = 0 or, C = \(\frac{I_1}{V_2}\) = 0.

143.

For the circuit given below, the value of the Transmission parameter B and D are __________(a) D = 0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω(b) D = 0.6923 + j0.5385 Ω, B = 6.923 + j25.385 Ω(c) D = -0.6923 + j0.5385 Ω, B= 25.385 + j6.923 Ω(d) D = -0.5385 + j0.6923 Ω, B = -6.923 + j25.385 ΩI had been asked this question in my homework.The doubt is from Advanced Problems Involving Parameters topic in section Two-Port Networks of Network Theory

Answer»

The correct choice is (a) D = 0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω

Easiest explanation: Z1 = \(\frac{(-j15)(-j10)}{-j15-j10-j20}\) = j10

Z2 = \(\frac{(-j10)(-j20)}{-j15}\) = \(-\frac{j40}{3}\)

Z3 = \(\frac{(j15)(-j20)}{-j15}\) = j20

-I2 = \(\frac{20-j40/3}{20-\frac{j40}{3}+j20}I_1 = \frac{3-j2}{3+j}\) I1

∴ D = \(\frac{-I_1}{I_2} = \frac{3+j}{3-j2}\) = 0.5385 + j0.6923 Ω

V1 = [j10 + 2(9+j7)] I1

= jI1 (24 – j18)

So, B = –\(\frac{V_1}{I_2}= \frac{-jI_1 (24-j18)}{-\frac{3-j2}{3+j} I_1}\)

= \(\frac{6}{13}\)(-15+j55)

∴ B = -6.923 + j25.385 Ω.

144.

For a T-network if the Short circuit admittance parameters are given as y11, y21, y12, y22, then y21 in terms of Inverse Hybrid parameters can be expressed as ________(a) y21 = \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)\)(b) y21 = \(\frac{g_{12}}{g_{22}} \)(c) y21 = –\(\frac{g_{21}}{g_{22}} \)(d) y21 = \(\frac{1}{g_{22}}\)I have been asked this question in an interview.My doubt stems from Inverse Hybrid (g) Parameter in portion Two-Port Networks of Network Theory

Answer»

The correct choice is (c) y21 = –\(\frac{g_{21}}{g_{22}} \)

The explanation is: We KNOW that, I1 = y11 V1 + y12 V2 ……… (1)

I2 = y21 V1 + y22 V2 ………. (2)

And, I1 = g11 V1 + g12 I2 ………. (3)

V2 = g21 V1 + g22 I2 ……….. (4)

Now, (3) and (4) can be rewritten as,

I1 = \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)V_1 + \frac{g_{12}}{g_{22}}V_2\)………. (5)

And I2 = –\(\frac{g_{21} V_1}{g_{22}} + \frac{V_2}{g_{22}}\)………. (6)

∴ Comparing (1), (2) and (5), (6), we get,

y11 = \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)\)

y12 = \(\frac{g_{12}}{g_{22}} \)

y21 = –\(\frac{g_{21}}{g_{22}} \)

y22 = \(\frac{1}{g_{22}}\).

145.

For the circuit given below, the value of Transmission parameter A and C are ____________(a) A = -0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω(b) A = 0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω(c) A = -0.7692 – j0.3461 Ω, C = -0.03461 + j0.023 Ω(d) A = 0.7692 – j0.3461 Ω, C = 0.023 + j0.03461 ΩThe question was asked by my college director while I was bunking the class.This is a very interesting question from Advanced Problems Involving Parameters topic in portion Two-Port Networks of Network Theory

Answer»

Right OPTION is (B) A = 0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω

To explain I would say: V = [20 + (-j10) || (j15 − j20)] I1

V1 = \(\BIG[20 + \frac{(-j10)(-j5)}{-j15}\Big]\) I1

= [20 – j\(\frac{10}{3}\)] I1

I0 = \(\left(\frac{-j10}{-j10-j5}\right)\) I1 = \(\frac{2}{3}\)I1

V2 = (-j20) I0 + 20I’0

= –\(\frac{j40}{3}I_1 + 20I_1 = (20 – \frac{j40}{3}) I_1 \)

∴ A = \(\frac{V_1}{V_2} = \frac{(20-\frac{j10}{3})I_1}{20-\frac{j40}{3}) I_1}\) = 0.7692 + j0.3461 Ω

∴ C = \(\frac{I_1}{V_2} = \frac{1}{20-j40/3}\) = 0.03461 + j0.023 Ω.

146.

For a T-network if the Short circuit admittance parameters are given as y11, y21, y12, y22, then y12 in terms of Inverse Hybrid parameters can be expressed as ________(a) y12 = \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)\)(b) y12 = \(\frac{g_{12}}{g_{22}} \)(c) y12 = –\(\frac{g_{21}}{g_{22}} \)(d) y12 = \(\frac{1}{g_{22}}\)I have been asked this question in an interview for job.I want to ask this question from Inverse Hybrid (g) Parameter in chapter Two-Port Networks of Network Theory

Answer»

The correct option is (b) y12 = \(\frac{g_{12}}{g_{22}} \)

Explanation: We know that, I1 = Y11 V1 + y12 V2 ……… (1)

I2 = y21 V1 + y22 V2 ………. (2)

And, I1 = g11 V1 + g12 I2 ………. (3)

V2 = g21 V1 + g22 I2 ……….. (4)

Now, (3) and (4) can be rewritten as,

I1 = \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\RIGHT)V_1 + \frac{g_{12}}{g_{22}}V_2\)………. (5)

And I2 = –\(\frac{g_{21} V_1}{g_{22}} + \frac{V_2}{g_{22}}\)………. (6)

COMPARING (1), (2) and (5), (6), we get,

y11 = \(\left(g_{11} – \frac{g_{12} g_{21}}{g_{22}}\right)\)

y12 = \(\frac{g_{12}}{g_{22}} \)

y21 = –\(\frac{g_{21}}{g_{22}} \)

y22 = \(\frac{1}{g_{22}}\).

147.

In two-port networks the parameter g22 is called _________(a) Short circuit input impedance(b) Short circuit current ratio(c) Open circuitvoltage ratio(d) Open circuit input admittanceThis question was posed to me in an interview for internship.This question is from Inverse Hybrid (g) Parameter topic in chapter Two-Port Networks of Network Theory

Answer»

Right choice is (b) Short circuit current ratio

To explain: We know that, G22 = \(\frac{V_2}{I_2}\), when V1 = 0.

Since the PRIMARY voltage terminal is short CIRCUITED and the ratio of the voltage and current in second LOOP is measured, THEREFORE the parameter g22 is called as Short circuit current ratio.

148.

In two-port networks the parameter g21 is called _________(a) Short circuit input impedance(b) Short circuit current ratio(c) Open circuitvoltage ratio(d) Open circuit input admittanceI had been asked this question in an interview.This intriguing question comes from Inverse Hybrid (g) Parameter topic in portion Two-Port Networks of Network Theory

Answer»

Right option is (c) Open circuitvoltage ratio

For explanation I would say: We know that, g21 = \(\frac{V_2}{V_1}\), when I2 = 0.

Since the SECOND output TERMINAL is short circuited when the ratio of the TWO VOLTAGES is measured, therefore the parameter g21 is called as Open circuit voltage ratio.

149.

In two-port networks the parameter g12 is called _________(a) Short circuit input impedance(b) Short circuit current gain(c) Open circuit reverse voltage gain(d) Open circuit output admittanceThis question was addressed to me in an interview for job.My doubt is from Inverse Hybrid (g) Parameter topic in section Two-Port Networks of Network Theory

Answer»

Right option is (c) Open CIRCUIT REVERSE voltage gain

Explanation: We know that, g12 = \(\frac{I_1}{I_2}\), when V1 = 0.

Since the primary terminal is short circuited and the ratio of the TWO currents is MEASURED, THEREFORE the parameter g12 is called as Short circuit current ratio.

150.

In two-port networks the parameter g11 is called _________(a) Short circuit input impedance(b) Short circuit current ratio(c) Open circuitvoltage ratio(d) Open circuit input admittanceI have been asked this question in exam.My question is taken from Inverse Hybrid (g) Parameter in portion Two-Port Networks of Network Theory

Answer»

The CORRECT answer is (d) Open CIRCUIT input admittance

Explanation: We know that, g11 = \(\frac{I_1}{V_1}\), when I2 = 0.

Since the second voltage TERMINAL is short CIRCUITED when the ratio of the current and voltage is measured, therefore the PARAMETER g11 is called as Open circuit input admittance.