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I have two questions I'm having trouble understanding. I greatly appreciate any help you all can give me! I'm not great at physics so the more details the better.

THANKS!!!

In the RC circuit of Figure WG32.6, the emf of the AC source is given by E = Emax sin ω| t. The emf amplitude is Emax = 10 V, and the angular frequency is ω,-100 s-1 when the angular frequency is changed to a new value ω2, the time average of the power delivered by the source decreases by a factor of 2. If R = 3.0 Ω and C = 5.0 × 10-2 F, what is the value of ω 2 DEVISE PLAN 2. What is the current amplitude as a function of o? 3. What is the phase difference as a function of o? 4·What is the time average of the power delivered by the AC source as a function of o? 5. Write an expression showing the ratio of the average power when ω and the average power when ω = ω 6. Solve this expression for 2. Figure WG32.6 3 EXECUTE PLAN 7. Determine ω, from the given values. 4 EVALUATE RESULT 8. Knowing that the average power decreases when the angular frequency changes from o to ω-, do you expect the amplitude of the current to increase or decrease? 1 GETTING STARTED 9. Does the average power vary as you expect as a function 1. Draw a phasor diagram for the circuit, showing Vk, Vc, an 10. Does your answer for ω2 agree with what you expect?

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Answer #1

QUESTIONS-2 :

(a) The amplitude of current will be given as :

using an ohm's law, we have

Imax = \varepsilonmax / R = (100 V) / (10 \Omega)

Imax = 10 A

And the phase constant will be given as :

using a formula, we have

\phi = tan-1 [XL - XC / R]

where, XL = inductive reactance = \omega L = (4000 s-1) (0.4 H)

XL = 1600 \Omega

XC = 1 / \omega C = 1 / (4000 s-1) (2 x 10-6 F)

XC = 125 \Omega

then, we get

\phi = tan-1 [(1600 \Omega) - (125 \Omega) / (10 \Omega)]

\phi = tan-1 (147.5)

\phi = 89.6 degree

(b) The ratio of the potential-difference amplitude across the inductor to the potential-difference amplitude across the capacitor which is given as :

VL / VC = I XL / I XC

VL / VC = (XL / XC) = (1600 \Omega) / (125 \Omega)

VL / VC = 12.8

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