Question

Consider a driven RLC circuit containing a resistance R = 100 Ohm, inductance L = 3 Times 10^-6 H, capacitance C = 1/3 Times 10^-10 F, and ac voltage source V = (0.002 V) cos(Ohm t), all in series. Carefully sketch the average power delivered vs Ohm, curve for this circuit, indicating all important features of the plot. P_ave^max = 2 Times 10^-8 W omega = 10^8 /s gamma = 1/3 Times 10^8 /s

Answer 1

Suppose a series RLC circuit is driven by a voltage source v()Vsint. Then the current i(t) that flows in the circuit is given by: () sin(+) in which: and (See the handout called The Series LRC Circuit for the details of the derivation of these results.) Because the voltage and current are functions of time, the power dissipated in the RLC circuit will also depend on time. The definition of the power dissipated in the RLC circuit at any time t is: The average power dissipated over one cycle is therefore given by: For the series RLC circuit, this becomes: So r(,)-blocos4||Γ.sin2 ω,dt|+/010 sín-, sinorcoser dt The integral in the second term is zero because the integrand sin at cosat is odd. Thus we have sin' ω t dt The integral in square brackets in Eq. (8) is the average value of sint over one cycle. This is equal to 1/2, as can be formally shown as follows. Use the trig identity

1-cos 2ot sin to write the integral in Eq. (8) as 2 2LTJo du To do the integral of cos 2 in Eq. (9), let -2at. Then du 2aodt and dtThen the integral 20 becomes cos u du = 2001, 0 The period T is related to the angular frequency ω by 2π so a)T-2t and 2a1T-4x . Therefore

ar, sinu -sin 4t-sin 0-0 Thus Eq. (9) becomes: and, from Eq. (8), the average power dissipated per cycle is: P(t)- 00 cos