A series RLC circuit has L = 0.01 mH, R = K ohm, C = 10 pF, in series with an ex

Question

A series RLC circuit has L = 0.01 mH, R = K ohm, C = 10 pF, in series with an external voltage source E (t) = 100 sin (omega t + pi/12) Volts. The switch is turned on At t = 0 seconds. a) Write the differential equation for the charge q on the capacitor. b) Find the value of omega so that the circuit is in resonance with the voltage source. c) For the value of omega found in part (a), find the form of the steady state current in the circuit. [Do not compute coefficients !] d) Find the amplitude of the steady-state current in the circuit for the value of omega found in part (a).

Explanation / Answer

8. Given series RLC circuit, L = 0.01 mH
R = 10 k ohm
C = 10 p F = 10 * 10^-12 F
E(t) = 100sin(wt + pi/12)
the switch is turned on at t = 0

a. so voltage across resistor = iR ( where i is the current in the circuit)
voltage across inductor = Ldi/dt
voltage across capacitor = q/C [ where dq/dt = i]
FROM KIRCHOFF'S voltage law
voltage drop across capacitor + voltage drop across resistor + voltage drop across inductor = net voltage
so, E(t) = iR + Ldi/dt + q/C
or, 100sin(wt + pi/2) = Rdq/dt + Ld^2q/dt^2 + q/c
b. for resonance w = wo = sqroot(1/LC) = sqroot(1/0.01*10^-3*10*10^-12) = 10^8 rad/s
c. during resonance, the imedance of the circuit becomes = R
so, E(t) = iR
i(t) = E(t)/R = 100sin(wt + pi/12)/10,000 = 0.01sin(10^8t + pi/12)
d. amplitude of this current = 0.01 A

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