PLEASE ANSWER THIS QUESTION, ITS URGENT....................
In a series LCR circuit with L = 5/3 H, R = 10 Ω and C = 1/30 F and a source of emf
E = 600 V, determine the charge on the capacitor as a function of time. It is given that
the initial charge on the capacitor is zero and the initial current in the circuit is 9A.
Answers
Given:
In a series LCR circuit with L = 5/3 H, R = 10 Ω and C = 1/30 F and a source of emf E = 600 V. It is given that the initial charge on the capacitor is zero and the initial current in the circuit is 9A.
To find:
Determine the charge on the capacitor as a function of time
Solution:
From given, we have,
In a series LCR circuit with L = 5/3 H, R = 10 Ω and C = 1/30 F and a source of emf E = 600 V.
The initial charge on the capacitor is zero
⇒ q (0) = 0
The initial current in the circuit is 9A
⇒ I (0) = 0 = q' (0)
The LCR circuit expression is given as follows.
L (d²q/dt²) + R (dq/dt) + (1/C) q = E
substitute the given values in the above equation, to further solve the above eqaution.
So, substitute, the inductance value, L = 5/3 H
The capacitance value, C = 1/30 F
The resistance value, R = 10 Ω
The emf of the source, E = 600 V
So, we get,
L (d²q/dt²) + R (dq/dt) + (1/C) q = E
5/3 (d²q/dt²) + 10 (dq/dt) + 30 q = 600
(d²q/dt²) + 6 (dq/dt) + 18 q = 360
Therefore, the solution of the IVP is as follows.
q(t) = e^{-3t} (20 e^{3t} - 20 cos (3t) - 17 sin (3t) )