Test: Second Order RLC Circuits- 2 - Electrical Engineering (EE) MCQ

v_R(t) = i(t) x 1 = -20 e^-2t = v_c(t) 

 = 80 e ^-2t
Now, i'(t) = i(t) + i_c(t)
= -20 e^-2t + 80 e^-2t = 60 e^-2t
So, 

or, 
v(t) = (60 - 30 - 20) e^-2t = 10 e^-2t

For an RLC series circuit to be critically damped

or, 

or, 120 R = 40 ( R + 120)
or, 3 R = R + 120 or R = 60 Ω

Given,

Taking Laplace transform on both sides,



or, 
or, 

Using partial fraction expansion,
A = 1, B = 1,
C = - 3

So, V(t) = (t + cost - 3 sint)

Given, 
= 1/a = 2
or, a = 0.5
Thus, 


or v(t) = 2(1 - e ^{- 0.5t})
Hence, v(1) = 2(1 - e^-0.5)

Characteristic equation of series RLC circuit is

Here, 

and 

or, 

For oscillatory response,

or, 

Applying KCL at the given node in s-domain, we have

or, 

or, 

Hence, there is no-zero and poles are at 4 s² + s + 2 = 0
or, 

Because the transient response of an electrical circuit can best be described by a differential equation hence, Laplace transformation finds its application in solving the transient behaviour of the electric circuits.

f(t) = e^-t sin 2t + e^-2t cos 5t
∴ 

∴ Find value of f(t) is

At t = 0⁺, capacitor will act as short circuit while inductor an open circuit

∴ V_20Ω = 10 volt

Applying node analysis,

or, 

or,

Here 

∴ 
or, 

At t = 0, inductor act as open circuit i(0⁺) = 0 A
At t = capacitor act as open circuit i(∞) = 0 A

At t = ∞, inductor will act as short circuit anc capacitor open circuit.

(characteristic equation)
Therefore 

If R = 0, ξ = 0 (vice-versa) and for ξ < 1, response will be oscillatory.