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The RLC Circuit
The RLC circuit is the electrical circuit consisting of a resistor of resistanceR, a coil of inductanceL, a capacitor of capacitanceCand a voltage source arranged in series. If the charge CR LV on the capacitor isQand the current flowing in the circuit isI, the voltage acrossR, LandCare RI, L dI dtandQCrespectively. By the Kirchhoff"s law that says that the voltage between any two points has to be independent of the path used to travel between the two points, LI ?(t) +RI(t) +1
CQ(t) =V(t)
Assuming thatR, L, CandVare known, this is still one differential equation in two unknowns,I andQ. However the two unknowns are related byI(t) =dQ dt(t) so that LQ ??(t) +RQ?(t) +1
CQ(t) =V(t)
or, differentiating with respect totand then subbing indQ dt(t) =I(t), LI ??(t) +RI?(t) +1
CI(t) =V?(t)
For an ac voltage source, choosing the origin of time so thatV(0) = 0,V(t) =E0sin(ωt) and the differential equation becomes LI ??(t) +RI?(t) +1
CI(t) =ωE0cos(ωt) (1)
The General Solution
We first guess one solution of (1) by tryingIp(t) =Asin(ωt-?) with the amplitudeAand phase?to be determined. That is, we are guessing that the circuit responds to an oscillating applied voltage with a current that oscillates with the same rate. ForIp(t) to be a solution, we need LI ??p(t) +RI?p(t) +1
CIp(t) =ωE0cos(ωt) (1p)
-Lω2Asin(ωt-?) +RωAcos(ωt-?) +1
CAsin(ωt-?) =ωE0cos(ωt)
=ωE0cos(ωt-?+?) and hence 1 C-Lω2?Asin(ωt-?) +RωAcos(ωt-?) =ωE0cos(?)cos(ωt-?)-ωE0sin(?)sin(ωt-?) c ?Joel Feldman. 2001. All rights reserved.1 Matching coefficients of sin(ωt-?) and cos(ωt-?) on the left and right hand sides gives ?Lω2-1
C?A=ωE0sin(?) (2)
RωA=ωE0cos(?) (3)
It is now easy to solve forAand?
(2) (3)=?tan(?) =Lω2-1 C
Rω=??= tan-1?LωR-1RCω?
(2)2+ (3)2=???Lω2-1 C?
2+R2ω2A=ωE0=?A=ωE0??Lω2-1
C?
2+R2ω2(4)
Naturally, different input frequenciesωgive different output amplitudesA. Here is a graph ofA againstω, with all other parameters held fixed. ωA Note that there is a small range of frequencies that give a large amplitude response. This is the phenomenon of resonance. It has been dramatically illustrated in, for example, the collapse of the
Tacoma narrows bridge.
Now back to finding the general solution. Note that subtracting (1p) from (1) gives
L(I-Ip)??(t) +R(I-Ip)?(t) +1
C(I-Ip)(t) = 0
That is, any solution of (1) differs fromIp(t) by a solution of LI ??(t) +RI?(t) +1
CI(t) = 0 (1c)
This is called the complementary homogeneous equation for (1). We now guess many solutions to (1 c) by tryingI(t) =ert, with the constantrto the determined. This guess is a solution of (1c) if and only if Lr
2ert+Rrert+1
Cert= 0??Lr2+Rr+1C= 0??r=-R±⎷
R2-4L/C
2L≡r1,2(5)
We now know thater1tander2tboth obey (1c). Because (1c) is linear and homogeneous, this forces c
1er1t+c2er2tto also be a solution, for any values of the constantsc1andc2. (To check this, just sub
c
1er1t+c2er2tinto (1c).) Assuming thatR2?= 4L/C,r1andr2are different and the general solution
to (1 c) isc1er1t+c2er2t.(It is reasonable to guess that, to solve a differential equation involving a second derivative, one has to integrate twice so that the general solution contains two arbitrary constants.) Then, the general solution of (1) is
I(t) =c1er1t+c2er2t+Asin(ωt-?)
withr1, r2given in (5) andA, ?given in (4). The arbitrary constantsc1andc2are determined by initial conditions. However, whener1tander2tdamp out quickly, as is often the case, their values are not very important. c ?Joel Feldman. 2001. All rights reserved.2quotesdbs_dbs2.pdfusesText_2
Chapter 21: RLC Circuits - Department of Physics