# Current peak time in RLC circuit?

1. Feb 17, 2014

### hobbs125

I am trying to figure out how to calculate the time at which current will peak in the following circuit.

Seems to me the capacitor will have no effect and the circuit will become a basic LR circuit in which the time constant is: 10/10010=.00099 x 5 = 4.9mS. Is this correct?

Any help will be greatly appreciated.

Last edited: Feb 17, 2014
2. Feb 18, 2014

### Staff: Mentor

Write the differential equation for the current, and solve it with the boundary condition of the switch closing at t=0.

3. Feb 19, 2014

If we use Laplace transform we can write the Laplace impedance of the capacity ZC=1/(s*C1) and R2 will be
Z12=ZC*R2/(ZC+R2)=R2/(s*C1)/(1/(s*C1)+R2)=R2/(1+R2*C1*s)
Then the i(s)=V1/s/[L1*s+R1+R2/(1+R2*C1*s)]
i(s)=V1*(1+R2*C1*s)/[ R2*C1*L1*s^3+(L1+R1*R2*C1)*s^2+(R1+R2)*s]
If R2*C1*L1*s^3+(L1+R1*R2*C1)*s^2+(R1+R2)*s=0 we get s3=0;s1,s2
Let’s put :
a= R2*C1*L1 ; b= (L1+R1*R2*C1); c=R1+R2
a=10000*5/10^9*10=0.0005; b=(10+10*10000*5/10^7)= 10.05; c=10+10000=10010.
Then a*s^2+b*s+c=0
s1=[-b+sqrt(b^2-4*a*c)]/(2*a) s2=[-b-sqrt(b^2-4*a*c)]/(2*a)
s1= -1056.787 ; s2= -18944.21
i(s)=V1*(1+R2*C1*s)/[a*(s-s1)*(s-s2)*s]=U(s)/[s*W(s)]
U(s)=V1*(1+R2*C1*s)/a ;
W(s)=(s-s2)*(s-s1)=s^2-(s1+s2)*s+s1*s2
W’(s)=2*s-(s1+s2)
Transforming in i(t):
i(t)=U(0)/W(0)+U(s1)/W’(s1)/s1*exp(s1*t)+ U(s2)/W’(s2)/s2*exp(s2*t)
Substituting s1 and s2 we get:
U(0)=V1/a =10/.0005 = 20000 ; W(0)=s1*s2= 20020000
U(s1)= 18943.21 ; U(s2)= 1055.787 ;W’(s1)= s1-s2 ; W’(s2)= s2-s1
The maximum it is t for di(t)/dt=0
di(t)/dt= U(s1)/W’(s1) *exp(s1*t)+ U(s2)/W’(s2) *exp(s2*t)=0
exp((s1-s2)*t)= U(s2)/W’(s2)*W’(s1)/U(s2)
tmax=ln(U(s1)/U(s2))/(s1-s2)=ln(18943.21 /1055.787)/( -1056.787 -( -18944.21))
tmax= 0.000161 sec

4. Feb 23, 2014

Sorry, I did not remark berkeman’s answer. It could be simpler, I guess.
i=total current; i1=capacitor current ; i2 = current through R2
V1=L1*di/dt+R1*i +R2*i2 first circuit
R2*di2/dt=-(L1*d2i/dt2+R1*di/dt)
integral(i1/C1*dt)=R2*i2 second circuit
i1/C1=R2*di2/dt ; i1=C1*R2*di2/dt; i1=-C1*(L1*d2i/dt2+R1*di/dt)
i=i1+ i2 i2=i-i1
i2=i+C1* (L1*d2i/dt2+R1*di/dt) then:
V1=L1*di/dt+R1*i +R2*[ i+C1* (L1*d2i/dt2+R1*di/dt) ] or:
R2*C1*L1*d2i/dt+(L1+R2*C1*R1)*di/dt+(R1+R2)*i=V1
a=R2*C1*L1 ; b=(L1+R2*C1*R1) ; c=(R1+R2)
ax^2+bx+c=0
x1=[-b+sqrt(b^2-4*a*c)]/(2*a)
x2=[-b-sqrt(b^2-4*a*c)]/(2*a)
x1= -1056.787 ; x2= -18944.21
i(t)=A*exp(x1*t)+B*exp(x2*t)+C
Since both x1 and x2 < 0 then for t=infinite C=V1/(R1+R2)
d2(i(t))/dt=x1^2*A*exp(x1*t)+x2^2*B* exp(x2*t)
If i(0)=0 then A+B+C=0 A+B=-C
If i(0)=0 then i2(0)=0 [at t=0]
i2=i+C1* (L1*d2i/dt2+R1*di/dt)
Then C1* (L1*d2i/dt2+R1*di/dt) =0
di/dt= x1*A*exp(x1*t)+x2*B*exp(x2*t)
d2i/dt2= x1^2*A*exp(x1*t)+x2^2*B*exp(x2*t) and
L1*d2i/dt2+R1*di/dt= L1*[A*x1^2+B*x2^2]+R1*[A*x1+B*x2]=0 or:
A*[L1*x1^2+R1*x1]+B*[L1*x2^2+R1*x2]=0 B/A=- (L1*x1^2+R1*x1)/( L1*x2^2+R1*x2)
B/A= -0.003109093
In order to find the maximum :
di/dt= x1*A*exp(x1*t)+x2*B*exp(x2*t)=0
x1*A*exp(x1*t)=-x2*B*exp(x2*t) -B/A=x1/x2*[exp(x2-x1)*t]
[exp(x2-x1)*t]=-B/A*x2/x1
tmax=ln(-x2/x1*B/A)/(x2-x1)=ln(18944.21/1056.787* 0.003109093) /( (18944.21-1056.787)= 0.000161 sec.

5. Feb 23, 2014

### Staff: Mentor

@Babadag -- your help is certainly appreciated, but it would help if you could format your replies in a more readable fashion. White space between paragraphs and equations helps a lot, and if you could post your equations in LaTeX, that would be even better (see the Feedback forum for a tutorial on LaTeX).

6. Feb 24, 2014