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.
A 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