به انگلیسی

Resonance and AC power

Science · Electricity & Magnetism · ages 20-22

Name ______________________ Date ____________

For sinusoidal voltage and current with phase difference phi, which formula gives average real power?
- Peak current times peak voltage
- Irms Vrms cos phi
- Irms Vrms divided by cos phi
What is the resonant angular frequency omega0 of an ideal series RLC circuit?
- 1 over sqrt(LC)
- sqrt(LC)
- R divided by L
An ideal inductor in sinusoidal steady state takes positive net energy from the supply over a full cycle.
Circle one: True False
A sinusoidal AC load has Irms = 1.5 A, Vrms = 12 V and cos phi = 0.8. What is its average real power?
- 18.0 W
- 14.4 W
- 22.5 W
An ideal series RLC circuit with R > 0 is driven at resonance. What is its impedance magnitude?
- Zero ohms
- Infinite ohms
- R ohms
Why does an ideal capacitor in sinusoidal steady state take zero net energy over a full cycle?
- It returns the field energy it stores
- It prevents any current from flowing
- It converts all stored energy to heat
An ideal LC-based tuner has fixed L. Its resonant frequency is twice the wanted frequency. How should its capacitance change to center resonance on the wanted frequency?
- Divide the capacitance by four
- Multiply the capacitance by four
- Multiply the capacitance by two
Two ideal series RLC circuits have the same L and C and equal fixed supply voltage amplitudes. One has smaller R. Compare peak current and half-power bandwidth.
- Smaller R: lower peak, narrower bandwidth
- Smaller R: higher peak, broader bandwidth
- Smaller R: higher peak, narrower bandwidth

Answer key

For grown-ups. Fold this page away before handing over the rest.

Irms Vrms cos phi · Average real power is the product of rms current, rms voltage and the power factor cos phi.
1 over sqrt(LC) · At resonance omega0 L = 1/(omega0 C). Therefore omega0^2 = 1/(LC), giving omega0 = 1/sqrt(LC).
False · An ideal inductor stores magnetic-field energy and returns it. Over a full periodic cycle, net energy taken is zero.
14.4 W · Pavg = 1.5 times 12 times 0.8 = 14.4 W. The rms current-voltage product alone is 18 VA.
R ohms · At resonance XL - Xc = 0. Thus Z = sqrt(R^2) = R because R is positive.
It returns the field energy it stores · An ideal capacitor stores electric-field energy and returns it. The amounts taken and returned balance over a full steady-state cycle.
Multiply the capacitance by four · At fixed L, f0 is proportional to 1/sqrt(C). Halving f0 requires doubling sqrt(C), so capacitance must be multiplied by four.
Smaller R: higher peak, narrower bandwidth · At resonance I0 = V0/R, so smaller R raises peak current. Angular half-power bandwidth is R/L, so smaller R narrows it.