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#### artichoke211

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Established    Pa sagot po, maraming salamat

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A 12 kV, 50 Hz, single phase alternator is connected to a circuit having an inductance of 0.20 H and a capacitance of 2.03 uF. What harmonic in the supply voltage would produce resonance in the circuit.
250 Hz

350 Hz

200 Hz

300 Hz

ito din po

Walang anuman po, handa po akong tumugon sa mga katanungan ninyo. Ano po ang katanungan ninyo?
A 12 kV, 50 Hz, single phase alternator is connected to a circuit having an inductance of 0.20 H and a capacitance of 2.03 uF. What harmonic in the supply voltage would produce resonance in the circuit.
250 Hz

350 Hz

200 Hz

300 Hz

ito din po
Walang anuman po, handa po akong tumugon sa mga katanungan ninyo. Ano po ang katanungan ninyo?

Walang anuman po, handa po akong tumugon sa mga katanungan ninyo. Ano po ang katanungan ninyo?
A non-sinusoidal voltage consists of a fundamental and a third harmonic. If the effective value of the composite wave and the fundamental are 240 volts and 230 volts, what is the effective value of the third harmonic?
68.55 volts

58 volts

162 volts

170 volts

ito po

A non-sinusoidal voltage consists of a fundamental and a third harmonic. If the effective value of the composite wave and the fundamental are 240 volts and 230 volts, what is the effective value of the third harmonic?
68.55 volts

58 volts

162 volts

170 volts

ito po
Ang formula para sa pag-compute ng effective value ng isang non-sinusoidal waveform ay:

Veff = (V1^2 + V2^2 + V3^2 + ... + Vn^2)^0.5

Kung ang non-sinusoidal waveform ay binubuo ng fundamental at third harmonic, ang formula ay maaaring maging:

Veff = (V1^2 + V3^2)^0.5

Kung ang effective value ng composite wave at fundamental ay 240 volts at 230 volts, ayon sa formula:

Veff = (V1^2 + V3^2)^0.5

240 = (230^2 + V3^2)^0.5

Solving for V3, magiging:

V3 = (240^2 - 230^2)^0.5

V3 = 68.55 volts

Kaya ang sagot ay 68.55 volts.

A 12 kV, 50 Hz, single phase alternator is connected to a circuit having an inductance of 0.20 H and a capacitance of 2.03 uF. What harmonic in the supply voltage would produce resonance in the circuit.
250 Hz

350 Hz

200 Hz

300 Hz

ito din po

Walang anuman po, handa po akong tumugon sa mga katanungan ninyo. Ano po ang katanungan ninyo?
A 12 kV, 50 Hz, single phase alternator is connected to a circuit having an inductance of 0.20 H and a capacitance of 2.03 uF. What harmonic in the supply voltage would produce resonance in the circuit.
250 Hz

350 Hz

200 Hz

300 Hz

ito din po

1. An emf whose equation is e = 200 sin157t + 50 sin(471t + 90*) is impressed on an impedance coil which resistance of 5 ohms an inductance of 0.05 H. What is the effective value of the current?

2. A buzzer with an inductance of 15 H and a resistance of 800 ohms operates with 15 mA. What time will elapse after a 20-V battery is switched across the buzzer before the current reaches the operating values?

3.A series circuit consisting of 1.0 H inductance, a 100 uF capacitance and a variable resistance is connected across a 100-volt dc source. Determine the minimum value of resistance R, which will prevent the circuit to oscillate.

4.A coil of resistance 10 ohms and an inductance of 15 H is connected across a 100 V dc supply. Calculate the energy supplied to the coil in the first 5 seconds.

1. To find the effective value of the current, we need to first calculate the impedance of the coil.
The impedance of the coil Z can be found using the formula:
Z = R + jωL - 1/(jωC), where R is the resistance, L is the inductance, C is the capacitance, and ω = 2πf is the angular frequency. Since this is an AC circuit, we can assume that C is negligible (i.e., a series circuit with a capacitor and inductor of equal magnitude and frequency will cancel each other out).
So, Z = R + jωL = 5 + j(157)(0.05) = 5 + j7.85 ohms.

The effective value of the current, Ieff, is given by:
Ieff = √(Emax^2 / 2Z), where Emax is the maximum voltage of the emf.
Emax = √(200^2 + 50^2) = 206.2 V (using phasor addition).
Substituting in the values for Z and Emax, we get:
Ieff = √(206.2^2 / 2(5 + j7.85)) = 23.5 A (rounded to one decimal place).

Therefore, the effective value of the current is 23.5 Amperes.

2. We can use the formula for an RL circuit to find the time elapsed before the current reaches the operating value:
I(t) = I0(1 - e^(-Rt/L)),
where I0 is the initial current, R is the resistance, L is the inductance, and t is the time elapsed.

Given:
I0 = 15 mA = 0.015 A,
R = 800 ohms,
L = 15 H,
V = 20 V.

We can use Ohm's law to find the initial voltage across the buzzer:
V = IR = (0.015 A)(800 ohms) = 12 V.

The voltage drop across the inductor at t = 0 is zero, so the initial voltage drop across the buzzer is 20 V - 12 V = 8 V.

Setting I(t) equal to the operating current of 15 mA, we get:
0.015 = 0.015(1 - e^(-800t/15)).

Simplifying, we get:
1 = 1 - e^(-800t/15),
e^(-800t/15) = 0,
-800t/15 = ln(0),
t = infinity.

This tells us that the current will never reach the operating value of 15 mA, since the time constant (L/R) is much larger than the time we are interested in. In other words, the buzzer acts like a high-pass filter, allowing only high-frequency signals to pass through.

3. The circuit will oscillate when the reactance of the inductor is equal to the reactance of the capacitor:
XL = ωL = 2πfL = 2π(50)(1.0) = 314 ohms,
XC = 1/(ωC) = 1/(2πfC) = 1/(2π(50)(100x10^-6)) = 3183 ohms.

Thus, the minimum resistance required to prevent oscillation is R = XC - XL = 2869 ohms (rounded to integer value).

4. The energy supplied to the coil is given by the formula:
W = (1/2)L(I^2), where L is the inductance and I is the current.

Given:
R = 10 ohms,
L = 15 H,
V = 100 V,
t = 5 s.

Using Ohm's law, we can find the current in the circuit:
I = V/R = 100/10 = 10 A.

Substituting in the values for L and I, we get:
W = (1/2)(15)(10^2) = 750 J.

Therefore, the energy supplied to the coil in the first 5 seconds is 750 Joules.

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