Question Solved1 Answer An RL circuit has an emf given (in volts) by 3 sin 2t, a resistance of 10 ohms, an inductance of 0.5 henry, and an initial current of 6 amperes. Find the current in the circuit at any time t.

ARF66M The Asker · Electrical Engineering
Transcribed Image Text: An RL circuit has an emf given (in volts) by 3 sin 2t, a resistance of 10 ohms, an inductance of 0.5 henry, and an initial current of 6 amperes. Find the current in the circuit at any time t.
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Transcribed Image Text: An RL circuit has an emf given (in volts) by 3 sin 2t, a resistance of 10 ohms, an inductance of 0.5 henry, and an initial current of 6 amperes. Find the current in the circuit at any time t.
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Step 1An RL Circuit{:[" emf "=3sin 2t],[R=10 Omega],[V=3sin 2t]:}Step 2Apply kirchhoff's routage Law{:[V=Ri(t)+L*(d^(i)(t))/(dt)],[=>L(di(t))/(dt)+Ri(t)=V],[=>(di(t))/(Delta t)+(R)/(L)i(t)=(V)/(L)],[=>(di(t))/(at)+(10)/(0.5)i(t)=(3sin 2t)/(0.5)],[=>(d^(')(t))/(Delta t)+20 i(t)=6sin 2t],[" Intergraiti "" Facter "I*F=e^(int_(20)20 dt)=e^(20 t)],[i(t)xx(I.F)=int6sin 2t(If)dt+A],[A rarr" Arbitary Consteant "],[I(t)e^(20 t)=int6sin 2t*e^(20 t)dt+A]:}I=int6sin 2t*e^(20 t)dtI=e^(20 t)int6sin 2tdt-int(d)/(at)(e^(20 t)g int6sin 2tdt]dtI=e^(20 t)*(b cos 2t)/(-2)-int(e^(20 t)*20 xx6cos 2t)/(-2)dtI=-3cos 2te^(20 t)+10 inte^(20 t)*6cos 2tdt=>I=-3cos 2t*l+60 inte^(20 t)cos 2tdt=>I=-3cos 2t*e^ ... See the full answer