# In the RLC circuit shown below, the passive element values are R=(13/3)O, L=(1/3)H, and C=(1/10)F, t

In the RLC circuit shown below, the passive element values are R=(13/3)Ω, L=(1/3)H, and C=(1/10)F, the voltage source is vs(t)=cos(50t)V. The initial voltage of the capacitor is known as vc(0)=6V. The open switch closes at t = 0, find the voltage in the capacitor, as a function of time, for t after the switch closes. As a “not to be submitted” exercise, do plot the current in the capacitor between t = 0 and t = 1 sec, and you’ll observe something extremely interesting: the steady state AC current has a small RMS value, but the initial transient clearly goes many times higher than that, thus endangering the equipment connected (Use the HP-50g plot for this).

(20 points) In the RLC circuit shown below, the passive element values are R = (13/3) Ω, L = (1/3) H. and C = (1/10) F, the voltage source is us (t) = cos (50) V. The initial voltage of the capacitor is known as v 0 6V The open switch closes at t = 0 find the voltage in the capacitor, as a function of time, for t after the switch closes. As a “not to be submitted” exercise, do plot the current in the capacitor between t0 and t 1 sec, and you&#39;ll observe something extremely interesting: the steady state AC current has a small RMS value, but the initial transient clearly goes many times higher than that, thus endangering the equipment connected (Use the HP-50g plot for this). Note: For the Heaviside function, use “u(t)”. Also, use e(x) or e*”(x) instead of exp(x) for the natural exponential function closes at t=0 Uc(t) = Note: Please enter your solution in the form (K1* e*(K2* t) + K3*e*(K4*t) all coefficients Ki. K5 cos(50t) + K6* sin(50t)) u(t) where you are to determine

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