×

Close

- Network Theory - NT
- Note
**Biju Patnaik University of Technology Rourkela Odisha - BPUT****13 Views**- Uploaded 10 months ago

NATIONAL INSTITUTE OF SCIENCE & TECHNOLOGY, PALUR HILLS, BERHAMPUR, ODISHA -761 008 CLASS TEST -2 SOLUTIONS Branch: ELECTRICAL ENGINEERING Semester : 3rd Subject with code : NETWORK THEORY (EE301T) Section : EEI-C Prepared by: Gubbala Kedarnath 1. Answer the all the questions 1 . i) ii) 1 X 10= 10 Marks A two port network is defined by the relations I1=2V1+V2, I2= 2V1+3V2. Then Z12is b) -1 Ω c) -1/2 Ω d) -1/4 Ω a) -2 Ω For the two port network shown in figure, the Zmatrix is given by a) Z1 Z + Z 2 1 Z1 + Z 2 Z 2 Z1 Z + Z 2 1 b) Z1 Z 2 c) Z1 Z 2 Z2 Z1 + Z 2 d) Z1 Z1 Z Z + Z 1 1 2 iii) The condition of reciprocity of a two port network having different parameters are iv) v) i) h12=-h21 ii) AD+BC =1 iii) Z12 =Z21 .choose the correct combination from the following 1 and 2 2 and 3 1,2, and 3 b) c) d) 1 and 3 a) The following property relates to LC impedance or admittance functions: a) The poles and zeros are simple and lie on the jω- axis b) There must be either a zero or a pole at origin and infinity c) The highest (or lowest) powers of numerator or denominator differ by unity. d) All of the above A reactance network in the First Foster form has a pole at ω=zero, and zero at ω=infinity. the element in the box 1 in network is a) An inductor A capacitor b) A parallel LC circuit c) d) A series LC circuit d) Reciprocal two port network: Y12=Y22 d) 0Ω vi) The parameter type and the matrix representation of the relevant two port parameters that describe the circuit shown are 0 0 h –parameters a) Z –parameters b) 0 0 1 0 0 1 0 0 1 0 Z –parameters c) h –parameters d) 0 0 0 1 vii) Which one of the following pairs is correctly matched? Symmetrical two Reciprocal two Symmetrical two b) port network: c) port network: a) port network: h11h22-h12h21=1 AD- BC=1 Z11=Z22 V (S ) viii) The transfer function 2 of the circuit shown V1 ( S ) below is a) b) 0.5s + 1 3s + 6 s +1 s+2 c) d) s+2 s +1 s +1 s+2 ix) The y- parameter y21 of the network shown in figure x) a) is 2mho b) is 6 mho c) is 3 mho d) does not exist Two networks are connected in cascade as shown in figure. With usual notations the equivalent A, B, C, D constants are obtained. Given that, C =0.025 ∠45 o , the value of Z is 2 a) 10∠30 o Ω b) 40∠ − 45 o Ω DEPARTMENT OF ELECTRICAL ENGINEERING c) 1Ω 2016-2017

NATIONAL INSTITUTE OF SCIENCE & TECHNOLOGY, PALUR HILLS, BERHAMPUR, ODISHA -761 008 Solutions 1i. From the given equations I1=2V1+V2, I2= 2V1+3V2. Compare with the Y- parameters governing equations I1=Y11V1+ Y12V2, I2=Y21V1+ Y22V2 Y12=1, ∆Y= 4 Z12 = − Y12 ∆Y = −1 4 Answer: D 1ii. Given network is KVL equations for the two loops V1=Z1I1+ Z1 I2, V2=Z1I1+ (Z1+ Z2)I2 Z11 Z 21 Z12 Z1 = Z 22 Z1 Z1 + Z 2 Z1 Answer: D 1iii. Answer: D 1iv. Answer: D 1v. According to the Foster First form realization If there is a pole at S=jω=0, the first element C0 is present If there is a pole at S=jω= ∞ , the last element Ln is present Consider this example Z ( s ) = (s 2 + 1)(s 2 + 9) s ( s 2 + 4) If, we substitute s=0 in Z(s),then Z(s)= ∞ . That is capacitor in s- domain will be 1 . So, it shows the property of a capacitor Z (s) = sC ∞ Similarly, If, we substitute s= ∞ ,in Z(s),then Z(s)= . So, ∞ DEPARTMENT OF ELECTRICAL ENGINEERING 2016-2017

NATIONAL INSTITUTE OF SCIENCE & TECHNOLOGY, PALUR HILLS, BERHAMPUR, ODISHA -761 008 s 2 (1 + 1 s Z (s) = 2 ) s 2 (1 + ss 2 (1 + 9 s 4 1 s (1 + ) s = 2 (1 + )(1 + 4 9 ) s 2 , Now substitute s= ∞ ) s s2 That is inductor in s- domain will be Z ( s) = sL . So, it shows the property of a inductor There is a pole at S=jω=0, the first element C0 is present 2 ) 2 There is a pole at S=jω= ∞ , the last element Ln is present. So, box 1 contains only series LC circuit. Answer: D 1vi. The given circuit diagram is From the circuit diagram V1=0, I2= 0 Z11 Z 21 Z12 0 ∞ undefined Y11 Y12 = = Z 22 Finite ∞ Y21 Y22 undefined Finite 0 ∞ h11 h12 0 0 A B 0 = = C D finite 0 h 21 h22 0 0 So, Answer :C 1vii. Answer :C 1viii. The given network is Apply KVL to the loop DEPARTMENT OF ELECTRICAL ENGINEERING 2016-2017

NATIONAL INSTITUTE OF SCIENCE & TECHNOLOGY, PALUR HILLS, BERHAMPUR, ODISHA -761 008 1 1 + (100 µ ) S (100 µ ) S 2 V1 ( S ) = 10k + (100 µ ) S S+2 V1 ( S ) = (100 µ ) S 1 S +1 = V2 ( S ) = 10k + (100 µ ) S (100 µ ) S V2 ( S ) S + 1 = V1 ( S )1 S + 2 V1 ( S ) = 10k + Answer: D 1ix. Given network is Apply KVL equations V1=12I1+6I2 and V2=20I1+10I2 Z11 Z12 12 6 Z = 21 Z 22 20 10 ∆Z = 0 − Z 21 Y21 = =∞ ∆Z So, Answer is does not exist. Answer: D 1x. The given network V1=AV2+BI2 I1=CV2+DI2 V2=Z2(I1+I2)………(i) DEPARTMENT OF ELECTRICAL ENGINEERING 2016-2017

## Leave your Comments