The capacity increases with the frequency

Sexl Physics 7, textbook

| 126 3 Basics of Electrical Engineering 1 A wire frame with an area of ​​100 cm 2 rotates in a homogeneous magnetic field (B = 0.05 T) at 50 revolutions per second. a) How big is the induced voltage for the positions 0 °, 45 °, 90 °, 135 °, 180 ° of the frame relative to the field? b) How many turns do you have to wind on the frame so that the maximum voltage is 10 V? 2 What is the capacitive resistance of a capacitor with a capacity of 1 µF for a) direct current? b) 50 Hz? c) 100 kHz? d) 100 MHz? Draw a graph for the capacitive resistance of a capacitor as a function of frequency. 3 A technical alternating voltage of U eff = 20 V (f = 50 Hz) is applied to a capacitor and a current of 100 mA is measured. How big is the capacity? 4 First, a DC voltage of 4 V is applied to a coil and a current of 0.1 A is measured. Then a technical alternating voltage of 12 V is applied and a current of 30 mA is measured. a) What is the ohmic resistance of the coil? b) What is the inductance of the coil? 5 On a toy transformer, which is intended for connection to a socket, you will find the inscription: Power P S = 30 W; maximum current I S = 2A. Calculate a) the secondary voltage U S, b) the ratio of the number of turns, c) the primary current I p. 6 Consider a medium power plant (P = 100 MW). The energy transfer should take place over a distance of 150 km with an aluminum line with a cross-section of A = 3 cm 2. Calculate the line resistance of the long-distance line from the total length l = 300 km of the outgoing and return lines and the specific resistance of aluminum, ρ = 3 · 10 –8 Ω m. With two different voltages (U = 110 kV, U = 380 kV), compare the power loss in the pipeline relative to the total output of 100 MW. What conclusion do you draw from your result? 5 Electromagnetic oscillations and waves 1 Calculate the wavelength of the radio program broadcast by Ö3 at the frequency 99.9 MHz. Find two more examples of known radio stations and calculate the wavelengths emitted from the associated frequencies. In which area of ​​the electromagnetic spectrum do these transmitters fall and what do you call the electromagnetic waves they emit? 2 An undamped electromagnetic resonant circuit contains a capacitor with a capacitance of 200 pF and a coil with an inductance of L = 10 -5 H. What is the natural frequency of the oscillating circuit? 3 The variable capacitor of a radio receiver has a variable capacitance of 50–500 pF. How large does the inductance of a coil connected in parallel have to be so that the entire medium wave range (0.5–1.5 MHz) can be received? 4 A resonant circuit with a natural frequency of 728 kHz is to be built from a capacitor with a capacitance of 365 pF and a coil. What inductance does the coil have to have? 5 By how many percent does the natural frequency of an oscillating circuit increase if its capacity is reduced by half? 6 The inductance of an oscillating circuit is 0.05 H, the capacitance 4 µF. Which sound is produced by this resonant circuit? 7 For a radio station there is an optimal antenna length depending on the broadcast frequency. Calculate the antenna length for the transmitter FM4 (103.8 MHz). 8 An electrical oscillating circuit has a resonance frequency f o. What is the resonance frequency in the following cases: a) In series with the coil, a second identical coil is built into the circuit. b) A second identical capacitor is installed in series with the capacitor. c) In parallel with the capacitor, a capacitor with eight times the capacity is installed. 9 If a second capacitor with a capacity of 33 nF is connected in series with the capacitor in an oscillating circuit, the resonance frequency increases by 15%. What is the capacity of the first capacitor? Solutions to the arithmetic problems 1 reflection and refraction 1 c = 2000 m / 1 s; d = 663 m 2 a) δ = 90 ° b) d = 2 15 cm tan β ≈ 19 cm 3 a) sin α / sin β = c 1 / c 2 = / b) λ 1 / λ 2 = c 1 / c 2 = / 4 B / G = b / g ¥ g = b * G / B = 0.09 m; from the lens equation follows f = 0.087 m 5 1 / g = 1 / f –1 / b = 1 / 0.05 m– (1 / - 0.15 m) ¥ g = 3.8 cm 6 c = 2 L / t = 4 L · N · Z = 313,000 km / s. 7 a) Displacement of the objective: 0.25 mm, 2.63 mm, 5.56 mm b) Magnification ratio: 5.025 · 10 -3; 5.263 x 10 -2; 0.1111 8 B / b = G / g = tan (viewing angle) = 0.0087. Since the object is very far away, b = f applies. a) image size B = 6.4 x 0.0087 mm = 0.056 mm; approx. 32 pixels wide. b) B = 32 x 0.0087 mm = 0.28 mm; approx. 160 pixels wide. c) In order for a crater to be recognizable, at least 3 × 3 pixels are necessary. A 7 km crater is 1/500 the diameter of the moon. At least 1500 pixels are therefore necessary for the full moon disk. You have to increase the focal length by a factor of 1500/160 to f = 300 mm. 9 a) f = 22 mm; b) f = 18 mm 10 If we insert the lens equation without glasses into the lens equation with glasses, we get: 1 / s –1 / g = 1 / f B ¥ g = 80 cm 2 The light wave 11 λ = 650 · 10 -9 m = 650 nm 12 red light φ 1 = 4 °, violet light φ 2 = 2 ° 18 '13 2 λ = d · sin φ ≈ d · tan φ = d · s / D ¥ d = 0.01 mm color red yellow green violet λ (air) λ (water) λ (glass) 700 nm 525 nm 467 nm 600 nm 450 nm 400 nm 500 nm 375 nm 333 nm 400 nm 300 nm 267 nm b) Thickness of the air layer: 2 d = 2k λ / 2 ¥ d = 300 nm thickness of the glass layer: 2 d = 2k λ G / 2 ¥ d = 200 nm 15 thickness of the soap lamella: d = λ W / 4 = λ / 4 n = 100 nm 16 a ) f = 1000 Hz b) (1) The radiation is perpendicular to the wall. (2) The radiation takes place in hemispherical waves, because the membrane can be viewed as a point source of sound. c) Low tones are heard better from the side than high tones. 17 D = 5 cm 18 a) (D / 2) 2 · π: (d / 2) 2 · π = 4 · 10 4 b) 1 “c) The resolution becomes better. 14 a) For test purposes only - property of the publisher öbv

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