1. A single core lead covered cable is to be designed for 132 kV to earth. Its conductor radius is 20 mm and its insulating material A, B and C have relative per mittivities 5, 4 and 3 respectively and corresponding maximum permissible stresses of 3.6, 2.4 and 2.0 kV/mm (rms) respectively. The minimum diameter of the lead steath will be _______ mm

1. 1. 173.78 mm

   
   
   

   2. 1738 mm

   
   
   

   3. 1.7378 mm

   
   
   

   4. 173.8 mm

   
   
   

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1. View Hint View Answer Discuss in Forum

   gmaxA =	q
   	2πε0εAr

   
   

   gmax B =	q
   	2πε0εBr1

   
   

   gmax C =	q
   	2πε0εCr2

   
   ε_Ar g_maxA = ε_Br₁g_{max B} = ε_Cr₂g_{max C}
   or 5 × 20 × 3.6 = 4 × r₁ × 2.4 = 3 × r₂ × 2.0
   

   ∴ r1 =	5 × 20 × 3.6	= 37.5 mm
   	4 × 2.4

   
   

   and r2 =	5 × 20 × 3.6	= 60 mm
   	3 × 2.0

   
   

   V1 gmax A r In		r1
   		r

   
   

   = 3.6 × 20 × In		37.5		= 45.25 kV
   		20

   
   

   V2 gmax B r1 In		r2
   		r1

   
   

   = 2.4 × 37.5 × In		60		= 42.3 kV
   		37.5

   
   

   V3 gmax C r2 In		R
   		r2

   
   

   = 2.0 × 60 × In		R
   		60

   
   

   Now, 132 = 45.25 + 42.30 + 120 ln		R
   		60

   
   

   ⇒ 44.45 = 120 ln		R
   		60

   
   

   ⇒ 0.370 = ln		R
   		60

   
   ⇒ = 86.89 mm
   ∴ Diameter of sheath = 2R = 2 × 86.89
   = 173.78 mm

   Correct Option: A

   gmaxA =	q
   	2πε0εAr

   
   

   gmax B =	q
   	2πε0εBr1

   
   

   gmax C =	q
   	2πε0εCr2

   
   ε_Ar g_maxA = ε_Br₁g_{max B} = ε_Cr₂g_{max C}
   or 5 × 20 × 3.6 = 4 × r₁ × 2.4 = 3 × r₂ × 2.0
   

   ∴ r1 =	5 × 20 × 3.6	= 37.5 mm
   	4 × 2.4

   
   

   and r2 =	5 × 20 × 3.6	= 60 mm
   	3 × 2.0

   
   

   V1 gmax A r In		r1
   		r

   
   

   = 3.6 × 20 × In		37.5		= 45.25 kV
   		20

   
   

   V2 gmax B r1 In		r2
   		r1

   
   

   = 2.4 × 37.5 × In		60		= 42.3 kV
   		37.5

   
   

   V3 gmax C r2 In		R
   		r2

   
   

   = 2.0 × 60 × In		R
   		60

   
   

   Now, 132 = 45.25 + 42.30 + 120 ln		R
   		60

   
   

   ⇒ 44.45 = 120 ln		R
   		60

   
   

   ⇒ 0.370 = ln		R
   		60

   
   ⇒ = 86.89 mm
   ∴ Diameter of sheath = 2R = 2 × 86.89
   = 173.78 mm

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2. A double circuit 1 – φ line is shown in the figure.
   
   Conductors a₁ and a₁ for ming one path are connected in parallel and carry the current in one direction; b₁ and b₂ forming the return path are connected in parallel and carry current in the one direction. If D₁ =1m, D₂ = 2m, the total inductance of the line per km will be _________ mH/km

1. 1. 12.6 mH/km

   
   
   

   2. 1.226 mH/km

   
   
   

   3. 12.26 mH/km

   
   
   

   4. 122.6 mH/km

   
   
   

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1. View Hint View Answer Discuss in Forum

   D_m = (D_a1b1 × D_a1b2 × D_a2b1 × D_a2b2)^1/4
   = (D₂ × √D²₁ + D²₂ × √D²₁ + D²₂ × D₂)^1/4
   (D₂ × √D²₁ + D²₂)^1/2
   D_sL = (D_a1b1 × D_a2b1 × D_a1b2)^1/4
   = 1r' × D₁ × D₁ × r')^1/4
   = (r' × D₁)^1/2
   Total inductance of the line is
   

   L = 2 ×		20 × 10–7 In		Dm
   				10

   
   

   = 4 × 10–7 In	(D2 × √D²1 + D²2)1/2
   	(r' × D1)1/2

   
   

   = 4 × 10–7 In	[2 × 51/2]1/2
   	(9.735 × 10–3)1/2

   
   = 2 × 10^–7(6.1298) H/m
   = 0.2 × 6.1298 mH/km
   = 1.226 mH/km

   Correct Option: B

   D_m = (D_a1b1 × D_a1b2 × D_a2b1 × D_a2b2)^1/4
   = (D₂ × √D²₁ + D²₂ × √D²₁ + D²₂ × D₂)^1/4
   (D₂ × √D²₁ + D²₂)^1/2
   D_sL = (D_a1b1 × D_a2b1 × D_a1b2)^1/4
   = 1r' × D₁ × D₁ × r')^1/4
   = (r' × D₁)^1/2
   Total inductance of the line is
   

   L = 2 ×		20 × 10–7 In		Dm
   				10

   
   

   = 4 × 10–7 In	(D2 × √D²1 + D²2)1/2
   	(r' × D1)1/2

   
   

   = 4 × 10–7 In	[2 × 51/2]1/2
   	(9.735 × 10–3)1/2

   
   = 2 × 10^–7(6.1298) H/m
   = 0.2 × 6.1298 mH/km
   = 1.226 mH/km

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3. A single circuit 50 Hz, 3 – φ transmission line has the following parameters per km.
   R = 0.2Ω, L = 1.3 mH, C = 0.01 µF
   The efficiency of the line, if the line is 120 km long and deliver 40 mW at 132 kV and 0.8 pf lagging will be _________%

1. 1. 0.926%

   
   
   

   2. 9.26%

   
   
   

   3. 92.6%

   
   
   

   4. 926%

   
   
   

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1. View Hint View Answer Discuss in Forum

   R = 0.2 × 120 = 24Ω
   X_L = 2πƒ. L = 314 × 120 × 1.30 × 10^-3
   = 48.98Ω
   X_C = 0.01 × 10^-6 × 120 = 12
   
   

   Ir =	40 × 1000	= 218.7 Amp
   	√3 × 132 × 0.8

   
   I_r = 218.7 (0.8 – j0.6) = (174.96 – j131.22)
   I_C1 = (j314)_r0.6 × 10^–6 × V_r
   

   Vr =	132 × 1000	= 76200 volts
   	√3

   
   I_C1 = (j314) × 0.6× 10^-6 × 76200 = (j14.356) amp
   I_L = I_C1 + I_r = (174.96 – j116.86)
   = 210.39 |33.73°
   loss = 3 × (210.39)² × 24 = 3.187 MW.
   

   η =	40 × 100	=	40 × 100	= 92.6%
   	(40 + 3.187		43.187

   Correct Option: C

   R = 0.2 × 120 = 24Ω
   X_L = 2πƒ. L = 314 × 120 × 1.30 × 10^-3
   = 48.98Ω
   X_C = 0.01 × 10^-6 × 120 = 12
   
   

   Ir =	40 × 1000	= 218.7 Amp
   	√3 × 132 × 0.8

   
   I_r = 218.7 (0.8 – j0.6) = (174.96 – j131.22)
   I_C1 = (j314)_r0.6 × 10^–6 × V_r
   

   Vr =	132 × 1000	= 76200 volts
   	√3

   
   I_C1 = (j314) × 0.6× 10^-6 × 76200 = (j14.356) amp
   I_L = I_C1 + I_r = (174.96 – j116.86)
   = 210.39 |33.73°
   loss = 3 × (210.39)² × 24 = 3.187 MW.
   

   η =	40 × 100	=	40 × 100	= 92.6%
   	(40 + 3.187		43.187

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4. The conductor of a 10 km long, single phase, two wire line are separated by a distance of 1.5 m. The diameter of each conductor is 1 cm. If the conductors are of copper, the inductance of the circuit is ________ mH

1. 1. 238

   
   
   

   2. 2.38

   
   
   

   3. 23.8

   
   
   

   4. 0.238

   
   
   

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1. View Hint View Answer Discuss in Forum

   Inductance of the circuit/km,
   

   L = 4 × 10–7 ln	D	mH/km
   	r'

   
   

   = 4 × 10–7 ln	1
   	0.7788 × 5 × 10–3

   
   = 2.381 mH/km.
   Inductance = 2.381 mH/km × 10 = 23.81 mH

   Correct Option: C

   Inductance of the circuit/km,
   

   L = 4 × 10–7 ln	D	mH/km
   	r'

   
   

   = 4 × 10–7 ln	1
   	0.7788 × 5 × 10–3

   
   = 2.381 mH/km.
   Inductance = 2.381 mH/km × 10 = 23.81 mH

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5. The surge impedance of a 400 km long overhead transmission line is 400 ohms. For a 200 km length of the same line, the impedance will be _______ Ω

1. 1. 0.4

   
   
   

   2. 400

   
   
   

   3. 0.04

   
   
   

   4. 40

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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