Answer to Question #344428 in Differential Equations for Zainab

Question #344428

Β An RCL circuit connected in series has R = 10 ohms, C = 10^2 farad, L = 1.5 henry, and an applied voltage E = 12 volts. Assuming no initial current and no initial charge at t = 0 when the voltage is first applied, find the subsequent current in the system.


1
Expert's answer
2022-05-30T03:35:06-0400
"Lq''+Rq'+\\dfrac{q}{C}=E"

"1.5q''+10q'+0.01q=12"

Homogeneous differential equation


"1.5q''+10q'+0.01q=0"

Corresponding (auxiliary) equation


"1.5r^2+10r+0.01=0"

"r=\\dfrac{-10\\pm\\sqrt{(10^2)-4(1.5)(0.01)}}{2(1.5)}"

"r_1=\\dfrac{-10-\\sqrt{99.94}}{3}"

"r_2=\\dfrac{-10+\\sqrt{99.94}}{3}"

The general solution of the homogeneous differential equation is


"q=c_1 e^{-(10+\\sqrt{99.94})t\/3}+c_2 e^{-(10-\\sqrt{99.94})t\/3}"

The general solution of the non homogeneous differential equation is


"q=c_1 e^{-(10+\\sqrt{99.94})t\/3}+c_2 e^{-(10-\\sqrt{99.94})t\/3}+1200"

"q(0)=0=>c_1+c_2+1200=0"

"c_2=-1200-c_1"

"i(t)=q'=-\\dfrac{10+\\sqrt{99.94}}{3}(c_1 e^{-(10+\\sqrt{99.94})t\/3})"

"-\\dfrac{10-\\sqrt{99.94}}{3}(c_2 e^{-(10-\\sqrt{99.94})t\/3})"

"i(0)=0"

"=>-\\dfrac{10+\\sqrt{99.94}}{3}c_1-\\dfrac{10-\\sqrt{99.94}}{3}c_2=0"


"-10c_1-\\sqrt{99.94}c_1+12000+10c_1"

"-1200\\sqrt{99.94}-\\sqrt{99.94}c_1=0"


"c_1=600(\\dfrac{10-\\sqrt{99.94}}{\\sqrt{99.94}})"

"c_2=-600(\\dfrac{10+\\sqrt{99.94}}{\\sqrt{99.94}})"

"i(t)=-\\dfrac{12}{\\sqrt{99.94}}e^{-(10+\\sqrt{99.94})t\/3}"

"+\\dfrac{12}{\\sqrt{99.94}} e^{-(10-\\sqrt{99.94})t\/3}"


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