Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Chapter 29, Problem 76P
To determine
To Show:Sharply peaked resonance,
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In a series RLC circuit, The source voltage Vs = 120Vac. The values of the components are: R1= 50.0 ohm, L1= 3.0 H, and C1= 6.0 x 10-6 F. If the angular frequency of the source equals the resonance frequency of the circuit, what is the average output power from the source?
RLC Circuit
=
Consider a series RLC circuit with L = 0.5 H, C = 3.0 μF, R
V(t) = Vo sin(wt) which has a maximum voltage amplitude Vo = 10V.
=
10.02 is connected to an AC voltage source
(a) What is the resonance frequency wo?
(b) Find the rms current(Irms) at resonance.
=
(c) Let apply a drive at frequency of w 3000 rad/s. Assume the current response is given by I(t) = I₁ sin(wt — 6)
Calculate the amplitude of the current(I) and the phase shift() between the current and the driving voltage.
In a series L-R-C ac circuit the current has the form
i(t) = 7 A cos(ω t)
while the ac source voltage is of the form
v(t) = 33 V cos(ω t + ϕ)
where ω = 377 rad/s and ϕ = −1 radians. Calculate (in Watts) the average power delivered to the circuit.
Chapter 29 Solutions
Physics for Scientists and Engineers, Vol. 1
Ch. 29 - Prob. 1PCh. 29 - Prob. 2PCh. 29 - Prob. 3PCh. 29 - Prob. 4PCh. 29 - Prob. 5PCh. 29 - Prob. 6PCh. 29 - Prob. 7PCh. 29 - Prob. 8PCh. 29 - Prob. 9PCh. 29 - Prob. 10P
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