In an oscillating series RLC circuit, find the timerequired for the maximum ener

Question

In an oscillating series RLC circuit, find the timerequired for the maximum energy present in the capacitor during anoscillation to fall to half its initial value. Assume q =Q at t = 0. (Use R, L,C, and Q, as necessary, where all quantities arein SI units.)

Explanation / Answer

An RLC circuit contains one or more resistors R, one or moreinductors L, and one or more capacitors C connected on a singleloop circuit. The loop rule produces the following equation L d/dt(dq/dt)+R dq/dt+q/C=0 The solution to this equation yields the following resultswhere =1/(L C) is the natural angular frequency of theoscillations '=(^2-(R/(2L))^2) is the quasi angular frequencyof the oscillations '˜ when >>R/(2L) which is true formost circuits f='/(2) is the frequency of the oscillations T=1/f=(2)/' is the period of the oscillations Q is the maximum charge or charge amplitude on the capacitor I is the maximum current or current amplitude through the circuitI=' Q ? is the phase angle of oscillations ?=arctan[i(initial)/('q(initial))] The energy stored in the Electric Field of the capacitor as afunction of time Uc(t)=(q(t)^2)/(2 C)=Q^2/(2C) e^(-R t/L)[cos(' t+ ?)]^2 The energy stored in the Magnetic Field of the inductor as afunction of time UL(t)=(L i(t)^2)/(2)=Q^2/(2C) e^(-R t/L)[sin(' t+ ?)]^2 The total energy stored in the circuit as a function of time Ut(t)=Uc(t)+UL(t)=Q^2/(2C) e^(-R t/L)[cos(' t+ ?)]^2+Q^2/(2C) e^(-R t/L) [sin(' t+?)]^2=Q^2/(2C) e^(-R t/L) Notice that all of the results above involve either e^(-R t/L)sin(' t+ ?) or e^(-R t/L) cos(' t+ ?) Functions of this type oscillate with an exponentially decreasingamplitude, meaning that their amplitudes trend to a value ofzero. Initially, when time t=0 e^(-R t/L) sin(' t+ ?)=sin(?) e^(-R t/L) cos(' t+ ?)=cos(?) which implies that all quantities start at a maximum value. After a long time, when time t e^(-R t/L) sin(' t+ ?)=0 e^(-R t/L) cos(' t+ ?)=0 which implies that all quantities trend to a minimum value ofzero. When energy stored in the Electric Field of the capacitorUc is a maximum equal to Q^2/2C e^(-R t/L),energy stored in the Magnetic Field of the inductor UL is a minimum equal to zero. When energy stored in the Magnetic Field of the inductorUL is a maximum equal to Q^2/2C e^(-R t/L),energy stored in the Electric Field of the capacitor Uc is a minimum equal to zero. The total energy of the circuit Ut decreaseswith time, and oscillates back and forth between The energy stored in the Electric Field of the capacitorUc The energy stored in the Magnetic Field of the inductorUL Now to solve the problem: The quantity we are interested here is the energy stored in theElectric Field of the capacitor as a function of time Uc(t)=(q(t)^2)/(2 C)=Q^2/(2C) e^(-R t/L)[cos(' t+ ?)]^2 This function has an amplitude maximum when cos(' t+ ?)=1 Uc max(t)=Q^2/(2C) e^(-R t/L) This function has an initial value when t=0 Uc max(0)=Q^2/(2C) e^(-R (0)/L)=Q^2/(2C)e^(0)=Q^2/(2C) This function has an arbitrary value at any time t Uc max(t)=Q^2/(2C) e^(-R t/L) We want to find the time when the value of Ucmax(t) at any time t is exactly half of the initial valueUc max(0). This is a special time known asthe half-life. We will set them equal with the 1/2 constant and solve for timet Uc max(t)=1/2 Ucmax(0) Q^2/(2C) e^(-R t/L)=1/2 (Q^2/(2C)) Divide both sides by Q^2/(2C) e^(-R t/L)=1/2 Take the natural log of both sides now ln(e^(-R t/L))=ln(1/2) -R t/L=ln(1/2) Solve for the time t -R t/L=ln(1/2) -R t/L=-0.693147 t=0.693147 L/R which can be input exactly as written here: 0.693147 L/R Notice that the answer does not depend at all on the capacitance C.All capacitors decay with the same half-life time.

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