Question #317101

Consider an electric circuit with an inductance of 5 henries

and a resistance of 12 ohms in series with an e.m.f 120sin200t

volts. Find the current


1
Expert's answer
2022-03-24T19:29:51-0400

Due to Kirchhoff’s law:

Ldidt+Ri=eL\frac {di}{dt}+Ri=e

5didt+12i=120sin200t5\frac {di}{dt}+12i=120\sin{200t}

The solutions to a nonhomogeneous equation are of the form

i(t)=ic(t)+ip(t)i(t) = i_c(t) + i_p(t)

where ic(t)i_c(t) is the general solution to the associated homogeneous equation and ip(t)i_p(t) is a particular solution.

The associated homogeneous equation:

5didt+12i=05\frac {di}{dt}+12i=0

ic(t)=Ce125ti_c(t)=Ce^{-\frac {12}{5}t}

ip(t)=Asin200t+Bcos200ti_p(t)=A\sin{200t}+B\cos{200t}

ip(t)=A200cos200tB200sin200ti_p’(t)=A\cdot200\cos{200t}-B\cdot200\sin{200t}

Now put these into the original differential equation to get:

5(A200cos200tB200sin200t)+5 (A\cdot200\cos{200t}-B\cdot200\sin{200t} )+ 12(Asin200t+Bcos200t)=120sin200t12(A\sin{200t}+B\cos{200t})=120\sin{200t}

A=9062509A=\frac{90}{62509} ; B=750062509B=-\frac{7500}{62509}.

Answer: i(t)=Ce125t+9062509sin200t750062509cos200ti( t)=Ce^{-\frac {12}{5}t}+ \frac{90}{62509}\sin{200t}-\frac{7500}{62509}\cos{200t}


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