Frequency of the electromagnetic wave,ν = 2.0 x 1010 Hz Electric field amplitude, Eo = 48 V m⁻¹ Speed of light, c = 3 x 108 m/s Ans (a). Wavelength of a wave is given by : λ = c/ν = (3 x 108) /(2.0 x 1010) = 0.015 m Ans (b). Magnetic field strength is given by: B0= E0/c = 48 /(3 x 108)= 1.6 x 10⁻⁷Read more
Frequency of the electromagnetic wave,ν = 2.0 x 1010 Hz
Electric field amplitude, Eo = 48 V m⁻¹
Speed of light, c = 3 x 108 m/s
Ans (a).
Wavelength of a wave is given by :
λ = c/ν
= (3 x 108) /(2.0 x 1010) = 0.015 m
Ans (b).
Magnetic field strength is given by:
B0= E0/c = 48 /(3 x 108)= 1.6 x 10⁻⁷ T
Ans (c).
Energy density of the electric field is given by :
UE = 1/2 ε0E²
And, energy density of the magnetic field is given as:
UB = 1/2 μ0B²
Where,
ε0 = Permittivity of free space
μ0 = Permeability of free space We have the relation connecting E and B as:
Energy of a photon is given as: E = hν = hc/λ Where, h = Planck’s constant = 6.6 x 10-34 Js c = Speed of light = 3 x 1o8 m/s λ = Wavelength of radiation Therefore, E = ( 6.6 x 10-34 x 3 x 1o8) /λ = (19.8 x 10⁻²⁶ )/λ J = (19.8 x 10⁻²⁶ )/( λ x 1.6 x 10⁻¹⁹) = (12.375x10⁻7 )/λ eV The given table lRead more
Energy of a photon is given as:
E = hν = hc/λ
Where,
h = Planck’s constant = 6.6 x 10-34 Js
c = Speed of light = 3 x 1o8 m/s
λ = Wavelength of radiation
Therefore,
E = ( 6.6 x 10-34 x 3 x 1o8) /λ
= (19.8 x 10⁻²⁶ )/λ J
= (19.8 x 10⁻²⁶ )/( λ x 1.6 x 10⁻¹⁹)
= (12.375×10⁻7 )/λ eV
The given table lists the photon energies for different parts of an electromagnetic spectrum for different λ are:
λ(m)
10³
1
10⁻³
10⁻⁶
10⁻⁸
10⁻¹⁰
10⁻¹²
E (eV)
12.375 x 10⁻¹⁰
12.375 x 10⁻⁷
12.375 x 10⁻⁴
12.375 x 10⁻¹
12.375 x 10¹
12.375 x 10³
12.375 x 10⁵
The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.
Electric field amplitude, Eo = 120 N/C Frequency of source, ν = 50.0 MHz = 50 x 106 Hz Speed of light, c = 3 x 108m/s Ans (a). Magnitude of magnetic field strength is given as: B0 =E0 /c = 120/(3 x 108) = 4 x 10⁻7 T= 400 nT Angular frequency of source is given as: ω =2πν = 2π x 50 x 106 = 3.14 x 108Read more
Electric field amplitude, Eo = 120 N/C
Frequency of source, ν = 50.0 MHz = 50 x 106 Hz
Speed of light, c = 3 x 108m/s
Ans (a).
Magnitude of magnetic field strength is given as:
B0 =E0 /c
= 120/(3 x 108)
= 4 x 10⁻7 T= 400 nT
Angular frequency of source is given as:
ω =2πν = 2π x 50 x 106 = 3.14 x 108 rad/s
Propagation constant is given as:
k = ω/c = (3.14 x 108 )/(3 x 108) =6.0 m
Ans (b).
Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular.
Amplitude of magnetic field of an electromagnetic wave in a vacuum, Bo = 510 nT = 510 x 10⁻9T Speed of light in a vacuum, c = 3 x 108 m/s Amplitude of electric field of the electromagnetic wave is given by the relation, E = cBo = 3 x 108x 510 x 10⁻9= 153 N/C Therefore, the electric field part of theRead more
Amplitude of magnetic field of an electromagnetic wave in a vacuum, Bo = 510 nT = 510 x 10⁻9T Speed of light in a vacuum, c = 3 x 108 m/s Amplitude of electric field of the electromagnetic wave is given by the relation, E = cBo
= 3 x 108x 510 x 10⁻9= 153 N/C Therefore, the electric field part of the wave is 153 N/C.
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.
A radio can tune to minimum frequency, ν1 = 7.5 MHz= 7.5 x 106 Hz Maximum frequency, ν2= 12 MHz = 12 x 106 Hz, Speed of light, c = 3 x 108 m/s Corresponding wavelength for ν1 can be calculated as: λ1=c/ν1 = (3 x 108)/( 7.5 x 106 ) =40m Corresponding wavelength for ν2 can be calculated as: λ2=c/ν2 =Read more
A radio can tune to minimum frequency,
ν1 = 7.5 MHz= 7.5 x 106 Hz Maximum frequency,
ν2= 12 MHz = 12 x 106 Hz,
Speed of light, c = 3 x 108 m/s
Corresponding wavelength for ν1 can be calculated as:
λ1=c/ν1 = (3 x 108)/( 7.5 x 106 ) =40m
Corresponding wavelength for ν2 can be calculated as:
λ2=c/ν2 = (3 x 108)/( 12 x 106 ) =25m
Thus, the wavelength band of the radio is 40 m to 25 m.
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular. Frequency of the wave, ν = 30 MHz = 30 x 106 s⁻1 Speed of light in a vacuum, c = 3 x 108 m/s Wavelength of a wave is given as:Read more
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular. Frequency of the wave, ν = 30 MHz = 30 x 106 s⁻1 Speed of light in a vacuum, c = 3 x 108 m/s Wavelength of a wave is given as: λ =c/ ν = (3 x 108)/(30 x 106) =10m
Radius of each circular plate, R = 6.0 cm = 0.06 m Capacitance of a parallel plate capacitor, C = 100 pF = 100 x 10-12 F Supply voltage, V = 230 V Angular frequency, ω = 300 rad s-1 Ans (a). Rms value of conduction current, I = v/ xc Where, Xc = Capacitive reactance =1/ωC Therefore ,I = V x ωC = 230Read more
Radius of each circular plate, R = 6.0 cm = 0.06 m Capacitance of a parallel plate capacitor, C = 100 pF = 100 x 10-12 F Supply voltage, V = 230 V Angular frequency, ω = 300 rad s–1 Ans (a).
Rms value of conduction current, I = v/ xc
Where, Xc = Capacitive reactance =1/ωC
Therefore ,I = V x ωC = 230 x 300 x 100 x 10–12 = 6.9 x 10-6 A = 6.9 μA
Hence, the rms value of conduction current is 6.9μA.
Ans (b).
Yes, conduction current is equal to displacement current.
Ans (c).
Magnetic field is given as: B = μ0rI0/2πR2
Where,
μ0 = Free space permeability = 4π x 10-7 NA–2
Io = Maximum value of current = √2I
r = Distance between the plates from the axis = 3.0 cm = 0.03 m
Therefore, B = (4π x 10-7 x 0.03 x √2 x 6.9 x 10-6) /(2π x 0.062)
=1.63 x 10-11 T
Hence, the magnetic field at that point is 1.63 x 10-11 T.
Radius of each circular plate, r = 12 cm = 0.12 m Distance between the plates, d = 5 cm = 0.05 m Charging current, I = 0.15 A Permittivity of free space, ε0 = 8.85 x 10⁻12 C2 N⁻1 m-2 Ans (a). Capacitance between the two plates is given by the relation, C = ε0 A/d Where, A = Area of each plate = πRead more
Radius of each circular plate, r = 12 cm = 0.12 m
Distance between the plates, d = 5 cm = 0.05 m
Charging current, I = 0.15 A
Permittivity of free space, ε0 = 8.85 x 10⁻12 C2 N⁻1 m-2
Ans (a).
Capacitance between the two plates is given by the relation,
C = ε0 A/d
Where, A = Area of each plate = πr2
C= ε0 πr2/d = [8.85 x 10⁻12 x π x (0.12)2 ]/0.05
= 8.0032 x 10⁻12 F = 80.032 pF
Charge on each plate, q = CV
Where,
V = Potential difference across the plates
Differentiation on both sides with respect to time (t) gives:
dq/dt = C dV/dt
But dq/dt = current (I)
Therefore ,dV/dt = I /C
=> 0.15/(8.0032 x 10⁻12) = 1.87 x 10⁹ V/s
Therefore, the change in potential difference between the plates is 1.87 x 109 V/s.
Ans (b).
The displacement current across the plates is the same as the conduction current.
Hence, the displacement current, id is 0.15 A.
Ans (c).
Yes
Kirchhoffs first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.
The rating of a step-down transformer is 40000 V-220 V. Input voltage,V, = 40000 V Output voltage,V2 = 220 V Total electric power required, P = 800 kW = 800 x 103 W Source potential, V = 220 V Voltage at which the electric plant generates power, V' = 440 V Distance between the town and power generatRead more
The rating of a step-down transformer is 40000 V-220 V. Input voltage,V, = 40000 V Output voltage,V2 = 220 V Total electric power required, P = 800 kW = 800 x 103 W Source potential, V = 220 V Voltage at which the electric plant generates power, V’ = 440 V Distance between the town and power generating station, d = 15 km
Resistance of the two wire lines carrying power =0.5 Ω/km Total resistance of the wire lines,
R = (15 + 15)0.5 = 15 Ω
P= V1I
Rms current in the wire line is given as:
=> I = P/V1
= 800 x 103/40000= 20 A
Ans (a).
Line power loss = I2R = (20)2 x 15 = 6 kW
Ans (b).
Assuming that the power loss is negligible due to the leakage of current.
Hence, power supplied by the plant = 800 kW + 6kW = 806 kW
Ans (c).
Voltage drop in the power line = IR = 20 x 15 = 300 V
Hence, voltage that is transmitted by the power plant
= 300 + 40000 = 40300 V
The power is being generated in the plant at 440 V.
Hence, the rating of the step-up transformer needed at the plant is 440 V – 40300 V.
Hence, power loss during transmission
= (600/1400 )x 100 = 42.8%
1400
In the previous exercise, the power loss due to the same reason is (6/806 )x 100 = 0.744%
Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 x 10¹⁰ Hz and amplitude 48 V m⁻¹. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field, [c = 3 x 10⁸m s⁻¹.]
Frequency of the electromagnetic wave,ν = 2.0 x 1010 Hz Electric field amplitude, Eo = 48 V m⁻¹ Speed of light, c = 3 x 108 m/s Ans (a). Wavelength of a wave is given by : λ = c/ν = (3 x 108) /(2.0 x 1010) = 0.015 m Ans (b). Magnetic field strength is given by: B0= E0/c = 48 /(3 x 108)= 1.6 x 10⁻⁷Read more
Frequency of the electromagnetic wave,ν = 2.0 x 1010 Hz
Electric field amplitude, Eo = 48 V m⁻¹
Speed of light, c = 3 x 108 m/s
Ans (a).
Wavelength of a wave is given by :
λ = c/ν
= (3 x 108) /(2.0 x 1010) = 0.015 m
Ans (b).
Magnetic field strength is given by:
B0= E0/c = 48 /(3 x 108)= 1.6 x 10⁻⁷ T
Ans (c).
Energy density of the electric field is given by :
UE = 1/2 ε0E²
And, energy density of the magnetic field is given as:
UB = 1/2 μ0B²
Where,
ε0 = Permittivity of free space
μ0 = Permeability of free space We have the relation connecting E and B as:
E = cB———————–Eq-1
Where,
c = 1/√(ε0μ0)————-Eq-2
Putting equation (2) in equation (1), we get
E = [1/√(ε0μ0)]B
Squaring both sides, we get
E² = [1/(ε0μ0)]B²
=> E²ε0 = B²/μ0
=>1/2 x E²ε0 = 1/2 x B²/μ0
=> UE = UB
See lessThe terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
Energy of a photon is given as: E = hν = hc/λ Where, h = Planck’s constant = 6.6 x 10-34 Js c = Speed of light = 3 x 1o8 m/s λ = Wavelength of radiation Therefore, E = ( 6.6 x 10-34 x 3 x 1o8) /λ = (19.8 x 10⁻²⁶ )/λ J = (19.8 x 10⁻²⁶ )/( λ x 1.6 x 10⁻¹⁹) = (12.375x10⁻7 )/λ eV The given table lRead more
Energy of a photon is given as:
E = hν = hc/λ
Where,
h = Planck’s constant = 6.6 x 10-34 Js
c = Speed of light = 3 x 1o8 m/s
λ = Wavelength of radiation
Therefore,
E = ( 6.6 x 10-34 x 3 x 1o8) /λ
= (19.8 x 10⁻²⁶ )/λ J
= (19.8 x 10⁻²⁶ )/( λ x 1.6 x 10⁻¹⁹)
= (12.375×10⁻7 )/λ eV
The given table lists the photon energies for different parts of an electromagnetic spectrum for different λ are:
The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.
See lessSuppose that the electric field amplitude of an electromagnetic wave is Eo = 120 N/C and that its frequency is v = 50.0 MHz. (a) Determine, Bo, ω, k, and λ. (b) Find expressions for E and B.
Electric field amplitude, Eo = 120 N/C Frequency of source, ν = 50.0 MHz = 50 x 106 Hz Speed of light, c = 3 x 108m/s Ans (a). Magnitude of magnetic field strength is given as: B0 =E0 /c = 120/(3 x 108) = 4 x 10⁻7 T= 400 nT Angular frequency of source is given as: ω =2πν = 2π x 50 x 106 = 3.14 x 108Read more
Electric field amplitude, Eo = 120 N/C
Frequency of source, ν = 50.0 MHz = 50 x 106 Hz
Speed of light, c = 3 x 108m/s
Ans (a).
Magnitude of magnetic field strength is given as:
B0 =E0 /c
= 120/(3 x 108)
= 4 x 10⁻7 T= 400 nT
Angular frequency of source is given as:
ω =2πν = 2π x 50 x 106 = 3.14 x 108 rad/s
Propagation constant is given as:
k = ω/c = (3.14 x 108 )/(3 x 108) =6.0 m
Ans (b).
Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular.
Equation of electric field vector is given as:
E = E0 sin (kx-ωt)j
= 120 sin [ 1.05 x -3.14 x 10⁸ t] j
And, Magnetic field vector is given as :
B = B0 sin (kx-ωt)k
= (4 x 10 sin⁻⁷ )[ 1.05 x -3.14 x 10⁸ t] k
See lessThe amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is Bo = 510 nT. What is the amplitude of the electric field part of the wave?
Amplitude of magnetic field of an electromagnetic wave in a vacuum, Bo = 510 nT = 510 x 10⁻9T Speed of light in a vacuum, c = 3 x 108 m/s Amplitude of electric field of the electromagnetic wave is given by the relation, E = cBo = 3 x 108x 510 x 10⁻9= 153 N/C Therefore, the electric field part of theRead more
Amplitude of magnetic field of an electromagnetic wave in a vacuum,
See lessBo = 510 nT = 510 x 10⁻9T
Speed of light in a vacuum, c = 3 x 108 m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cBo
= 3 x 108x 510 x 10⁻9= 153 N/C
Therefore, the electric field part of the wave is 153 N/C.
A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.
See lessA radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
A radio can tune to minimum frequency, ν1 = 7.5 MHz= 7.5 x 106 Hz Maximum frequency, ν2= 12 MHz = 12 x 106 Hz, Speed of light, c = 3 x 108 m/s Corresponding wavelength for ν1 can be calculated as: λ1=c/ν1 = (3 x 108)/( 7.5 x 106 ) =40m Corresponding wavelength for ν2 can be calculated as: λ2=c/ν2 =Read more
A radio can tune to minimum frequency,
ν1 = 7.5 MHz= 7.5 x 106 Hz Maximum frequency,
ν2= 12 MHz = 12 x 106 Hz,
Speed of light, c = 3 x 108 m/s
Corresponding wavelength for ν1 can be calculated as:
λ1=c/ν1 = (3 x 108)/( 7.5 x 106 ) =40m
Corresponding wavelength for ν2 can be calculated as:
λ2=c/ν2 = (3 x 108)/( 12 x 106 ) =25m
Thus, the wavelength band of the radio is 40 m to 25 m.
See lessA plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular. Frequency of the wave, ν = 30 MHz = 30 x 106 s⁻1 Speed of light in a vacuum, c = 3 x 108 m/s Wavelength of a wave is given as:Read more
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular. Frequency of the wave, ν = 30 MHz = 30 x 106 s⁻1
See lessSpeed of light in a vacuum, c = 3 x 108 m/s Wavelength of a wave is given as:
λ =c/ ν = (3 x 108)/(30 x 106) =10m
A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C=100pF.The capacitor is connected to a 230 V ac supply with a (angular) frequency of to a 230 V ac supply with a (angular) frequency of 300 rad s⁻¹. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
Radius of each circular plate, R = 6.0 cm = 0.06 m Capacitance of a parallel plate capacitor, C = 100 pF = 100 x 10-12 F Supply voltage, V = 230 V Angular frequency, ω = 300 rad s-1 Ans (a). Rms value of conduction current, I = v/ xc Where, Xc = Capacitive reactance =1/ωC Therefore ,I = V x ωC = 230Read more
Radius of each circular plate, R = 6.0 cm = 0.06 m
Capacitance of a parallel plate capacitor, C = 100 pF = 100 x 10-12 F Supply voltage, V = 230 V
Angular frequency, ω = 300 rad s–1
Ans (a).
Rms value of conduction current, I = v/ xc
Where, Xc = Capacitive reactance =1/ωC
Therefore ,I = V x ωC = 230 x 300 x 100 x 10–12 = 6.9 x 10-6 A = 6.9 μA
Hence, the rms value of conduction current is 6.9μA.
Ans (b).
Yes, conduction current is equal to displacement current.
Ans (c).
Magnetic field is given as: B = μ0rI0/2πR2
Where,
μ0 = Free space permeability = 4π x 10-7 NA–2
Io = Maximum value of current = √2I
r = Distance between the plates from the axis = 3.0 cm = 0.03 m
Therefore, B = (4π x 10-7 x 0.03 x √2 x 6.9 x 10-6) /(2π x 0.062)
=1.63 x 10-11 T
Hence, the magnetic field at that point is 1.63 x 10-11 T.
See lessFigure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of charge of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoffs first rule (junction rule) valid at each plate of the capacitor? Explain.
Radius of each circular plate, r = 12 cm = 0.12 m Distance between the plates, d = 5 cm = 0.05 m Charging current, I = 0.15 A Permittivity of free space, ε0 = 8.85 x 10⁻12 C2 N⁻1 m-2 Ans (a). Capacitance between the two plates is given by the relation, C = ε0 A/d Where, A = Area of each plate = πRead more
Radius of each circular plate, r = 12 cm = 0.12 m
Distance between the plates, d = 5 cm = 0.05 m
Charging current, I = 0.15 A
Permittivity of free space, ε0 = 8.85 x 10⁻12 C2 N⁻1 m-2
Ans (a).
Capacitance between the two plates is given by the relation,
C = ε0 A/d
Where, A = Area of each plate = πr2
C= ε0 πr2/d = [8.85 x 10⁻12 x π x (0.12)2 ]/0.05
= 8.0032 x 10⁻12 F = 80.032 pF
Charge on each plate, q = CV
Where,
V = Potential difference across the plates
Differentiation on both sides with respect to time (t) gives:
dq/dt = C dV/dt
But dq/dt = current (I)
Therefore ,dV/dt = I /C
=> 0.15/(8.0032 x 10⁻12) = 1.87 x 10⁹ V/s
Therefore, the change in potential difference between the plates is 1.87 x 109 V/s.
Ans (b).
The displacement current across the plates is the same as the conduction current.
Hence, the displacement current, id is 0.15 A.
Ans (c).
Yes
Kirchhoffs first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.
See lessDo the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?
The rating of a step-down transformer is 40000 V-220 V. Input voltage,V, = 40000 V Output voltage,V2 = 220 V Total electric power required, P = 800 kW = 800 x 103 W Source potential, V = 220 V Voltage at which the electric plant generates power, V' = 440 V Distance between the town and power generatRead more
The rating of a step-down transformer is 40000 V-220 V.
Input voltage,V, = 40000 V
Output voltage,V2 = 220 V
Total electric power required,
P = 800 kW = 800 x 103 W
Source potential, V = 220 V
Voltage at which the electric plant generates power,
V’ = 440 V
Distance between the town and power generating station, d = 15 km
Resistance of the two wire lines carrying power =0.5 Ω/km Total resistance of the wire lines,
R = (15 + 15)0.5 = 15 Ω
P= V1I
Rms current in the wire line is given as:
=> I = P/V1
= 800 x 103/40000= 20 A
Ans (a).
Line power loss = I2R = (20)2 x 15 = 6 kW
Ans (b).
Assuming that the power loss is negligible due to the leakage of current.
Hence, power supplied by the plant = 800 kW + 6kW = 806 kW
Ans (c).
Voltage drop in the power line = IR = 20 x 15 = 300 V
Hence, voltage that is transmitted by the power plant
= 300 + 40000 = 40300 V
The power is being generated in the plant at 440 V.
Hence, the rating of the step-up transformer needed at the plant is 440 V – 40300 V.
Hence, power loss during transmission
= (600/1400 )x 100 = 42.8%
1400
In the previous exercise, the power loss due to the same reason is (6/806 )x 100 = 0.744%
Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.
See less