General RLC solution Example 9 by Neil Cotter, Exams of Electrical Engineering

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By: Neil E. Cotter RLC CIRCUITS GENERAL RLC SOLUTION Example 9 Ex: 1 1024 -¥—_——————_ + v(t) - 2 250mH 130k&2 20 kQ 40v In the above circuit, the switch moves from a to b att = 0. Given that no energy is stored in the inductor at time t= 0, find a numerical expression for v(t) for t > 0. Solw: First, we find the initial conditions, t, (07) and wg Com), Since there is no energy shored in the inductor and energy w= LLTLO), Zz we have i, 7) = OA. The capacitor will charge to 40V measured with ++ on the Left gide of C: v, (07) = ov. Sinee the inductor current and capaci-tor voltage ao not change instantly, we have iL Cot) = i,(o7) 2 0A, velot)= velo) = 4oV Second, we caleulate the characteristic roots of the circuit for +>0,