Question

An (open) electrical circuit consists of an inductor, a resistor, and a capacitor. There is an initial charge of 1 coulomb on the capacitor. At the instant the circuit is closed, a current of 5 amperes is present and a voltage of E(t) = 23 cost is applied. In this circuit the voltage drop across the resistor is 5 times the instantaneous change in the charge, the voltage drop across capacitor is 11 times the charge, and the voltage drop across the inductor is 2 times the instantaneous change in the current. Write an initial value problem to model the circuit. Write the differential equation that models this circuit in terms of the charge, q. Write the initial conditions for the given circuit. 960) = (simplify your answer.) q'(0)=(Simplify your answer.)

Answer 1

Here given that Elt)= 23 cost Voltage drop accross R is sXdq at Voltage drop is 1x q Voltage drop across the inductor is 2x d²a accross C at2 t=0 L at t=0 210) q'LO) = 5 dt utsing Kirchhoff's law L d²q + R da ti q Elt) at 2 at Now 1 c z => 5 d'a t 5dq til di² at q 23 Cost 59" I 59' +119 23 Cost Q10) = 1 q 970) = 5 CS