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franciscovazquez623
04.08.2021 •
Physics
An alternating voltage of 100V, 50HZ Is Applied across an impedance of (20-j30) calculate the resistance, the capacitance, current, the phase angle between current and voltage
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Ответ:
The resistance R = 20 Ω
The capacitance C = 106.1 μF
The current, I is 2.773 A at 56.31°.
The phase angle of the between the current and the voltage is 56.31° leading.
Since the impedance Z = 20 - j30 Ω, the resistance, R is the real part of the impedance. So R = ReZ = 20 Ω
So, the resistance R = 20 Ω
To find the capacitance, we need first to find the reactance of the capacitor X. Since the impedance Z = 20 - j30, the reactance of the capacitor X. is the imaginary part of the impedance. So X = ImZ = 30 Ω.
Now the reactance of the capacitor X = 1/ωC where ω = angular frequency of the circuit = 2πf where f = frequency of the circuit = 50 Hz and C = capacitance
So, C = 1/ωX = 1/2πfX
Substituting the values of the variables into the equation, we have
C = 1/2πfX
C = 1/(2π × 50 Hz × 30 Ω)
C = 1/3000π
C = 1/9424.778
C = 1.061 × 10⁻⁴ F
C = 106.1 × 10⁻⁶ F
C = 106.1 μF
So, the capacitance is 106.1 μF
The current I = V/Z where V = voltage = 100 V at 0° and Z = impedance.
The magnitude of Z = √(20² + (-30)²)
= √(400 + 900)
= √1300
= 36.06 Ω
and its angle Φ = tan⁻¹(ImZ/ReZ)
= tan⁻¹(-30/20)
= tan⁻¹(-1.5) = -56.31°
So, V = 100 ∠ 0° and Z = 36.06 ∠ -56.31°
So, the current, I = V/Z = (100 ∠ 0°)/36.06 ∠ -56.31°
= 100/36.06 ∠(0° - (-56.31° ))
= 2.773 ∠ 56.31° A
So, the current is 2.773 A at 56.31°.
Since the current is 2.773 A at 56.31°, the phase angle of the between the current and the voltage is 56.31° leading.
So, the phase angle of the between the current and the voltage is 56.31° leading.
Learn more about alternating voltage here:
Ответ:
The more energy a wave carries the higher its amplitude.
Explanation:
The energy of a wave is defined by the equation:
Where
is energy,
is mass,
is angular velocity, and
is amplitude.
The previous equation indicates that the energy is directly proportional to the amplitude; if the energy of a wave increases so does the amplitude, and if the energy of a wave decreases the amplitude also decreases.
Thus the answer is:
The more energy a wave carries the higher its amplitude.