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The circuit in Figure P7-10 is in the zero state. Find the voltage v0(t) for t>0 when an input of is(t) = Iau(t) is applied. Identify the forced and natural components in the output.

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|p-1|{:[i_(S)(t)=I_(A)u(t)],[" For "t < 0","quadi_(S)(t)=I_(A)*0=0" Amp "],[" for, "t > 0" is "{:[S(t) > I_(A)*1=I_(A)A_(mp)],[" as we know "u(t)=1" fw "t > 0],[=0" fw "t < 0]:}]:}For t < 0,quadi_(s)(t)=0AmpThe Circmit Becomes[at steady stateCapacitor replaced by open cirmir] is opencircmits.N_(C)(oo)=R*I_(A)For Finding the time constant of the ciranit open the current sourec'Scanned withCamScanner|p-2|The V_(0)(t) at t >= 0v(t)=v(oo)+[v(0)-v(oo)]e^(-t//tau)if we replueed als the above Calculated Value and putit into the above equation we got the v_(0)(t) at t >= 0.By previous calculation we gor{:[V(0)=0V_(0)t],[V(oo)=R*I_(A)],[tau" (time const ")=RC],[:.V(t) > R*I_(A)+[0-R*I_(A)]e^(-t//RC)],[=R*I_(A)-RI_(A)e^(-t//RC)],[V ... See the full answer