An RLcircuit with a 5 resistor and a 0.05-H inductor carries a current of 1 A at t= 0, at which time a voltage source E(t) = 5 cos 120tV is added. Determine the subsequent inductor current and voltage.

Short Answer

Expert verified
  • The subsequent inductor current isI(t)=cos(120t)+1.2sin(120t)+1.44e-100t2.44.
  • The subsequent inductor voltage EL(t)=-6sin(120t)+7.2cos(120t)-7.2e-100t2.44

Step by step solution

01

Determine the subsequent inductor current

Here given R=5Ω,L=0.05H,It=1,Et=5cos(120t)V

Apply

I(t)=e-RtLeRtLE(t)Ldt+KI(t)=e-5t0.05e5t0.055cos(120t)0.05dt+KI(t)=e-100te100tcos(120t)dt+K

02

Solving the integral part ∫e100tcos(120t)dt.

e100tcos(120t)dt=e100tcos(120t)100+120sin(120t)e100t100=e100tcos(120t)100+65sin(120t)e100tdt=e100tcos(120t)100+65e100tsin(120t)100-120cos(120t)e100t100dte100tcos(120t)100+65e100tsin(120t)100-1.44cos(120t)e100tdt

Put A=e100tcos(120t)dt

role="math" localid="1664219728153" A=e100tcos(120t)100+65e100tsin(120t)100-1.44AA+1.44A=e100tcos(120t)100+65e100tsin(120t)1002.44A=e100tcos(120t)100+65e100tsin(120t)100A=e100tcos(120t)+1.2e100tsin(120t)244

I(t)=e-100t100(e100tcos(120t)+1.2e100tsin(120t)244)+KI(t)=cos(120t)+1.2sin(120t)2.44+Ke-100t

Put I(0) = 1 ,then K = 1.44/2.44

I(t)=cos(120t)+1.2sin(120t)2.44+1.442.44e-100tI(t)=cos(120t)+1.2sin(120t)+1.44e-100t2.44

Hence, the subsequent inductor current isI(t)=cos(120t)+1.2sin(120t)+1.44e-100t2.44.

03

Evaluate the subsequent indicator voltage

EL(t)=LdIdt=0.05ddtcos(120t)+1.2sin(120t)+1.44e-100t2.44=0.05-120sin(120t)+7.2cos(120t)-7.2e-100t2.44=-6sin(120t)+7.2cos(120t)-7.2e-100t2.44

Hence, the subsequent inductor voltage EL(t)=-6sin(120t)+7.2cos(120t)-7.2e-100t2.44

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