Question

1)

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 7 amperes is present and a voltage of E(t) = 20 cos t is applied. In this circuit the voltage drop across the resistor is 7 times the instantaneous change in the​ charge, the voltage drop across the capacitor is 11 times the​ charge, and the voltage drop across the inductor is 5 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.

? q'' + ? q' + ? q = ?

Write the initial conditions for the given circuit.

q(0) = ?

q'(0 )= ?

Answer 1

Given initial charge Let q (t) be instantaneous charge , i(t) = da IC be ing. aurent. q(o)=1c initial Current i(o)=7A or q co) = 7A so dt Voltage drop across incluctor is 5 di Ve as al de la sais Voltage Drop across Resistance = 7 dq = VR dt R C mums | Votedage voltage Prop across Con Capacitor 119=Vc Ve Elt) = 20cost By Applying Kirchoff's Law To circuit, V+ Vrt v = v= E(T) sd²a +7 dat 119=20cost dt² dt So Diff. es. is 59"+79 +119= 20 cost, G's) :74. 970) a IC