Test: Step Response of First Order Circuits - 2 - Electrical Engineering (EE) MCQ

At t ≤ 0, V_c (0) = 0 V

Time constant = τ = R_eq C_eq

R_eq = 2kΩ || 3 kΩ = 1.2 kΩ

C_eq = 5 nF

τ = RC = 6 μsec

At t = 6 μsec

⇒ V_c(t) = 3 [1 - e^-1] = 1.896 V.

Concept: 

We can use Kirchhoff’s Voltage Law, (KVL) to define the individual voltage drops that exist around the circuit and then hopefully use it to give us an expression for the flow of current.

The Time Constant, ( τ ) of the LR series circuit is given as L/R and in which V/R represents the final steady-state current value after five-time constant values.

Calculation:

L = 1 H, R = 10Ω

Time constant = 1/10 = 0.1 sec

First order system is defined by total number of poles and also which is same as the order of differential equation.

G(s) = 100/(s + 1) (s + 100)
Taking the dominant pole consideration,
S = -100 pole is not taken.
G(s) = 100/s + 1
Now it is first order system, ts  4T = 4 sec.

time constant = 0.2 sec.
1/a = 0.2
a = 5
final value =
K/a = 2
K = 10.

Concept:

Source voltage RMS value of a series RL circuit will be;

V_rms = √(V_R² + V_L²)

Where;

V_rms → Source voltage in RMS.

VR  → Voltage across resistance in RMS.

V_L  → Voltage across inductor in RMS.

RMS voltage = Peak voltage/√(2) 

Calculation:

Given;

VR = 13 V

VL  = 12 V

Vrms = √(VR2 + VL2) = √(13² + 12²) = 17.69

Peak value of the source voltage = Vrms × √2 = 25.01 V ≈ 25 V

Series RL circuit:

Let the voltage be V_s(t) = Vm sin ωt

The transient current equation is 

where

V_m = Peak value of voltage

θ = Peak value of impedance angle

ω = angular frequency

R = Resistance

L = Inductance

α = instant at which the circuit is closed

θ = Impedance angle

The exponential decay term represents the transient term and the remaining is the steady-state term.

The response is shown below;

In the steady-state, the RLC circuit elements give a response that is in synchronization to the input frequency.

Because transient analysis expressions consist of exponential decay terms.

Size of transient current primarily depends on the instant in the voltage cycle at which circuit is closed i.e. it depends on α value.

Concept:

Calculation:

Taking Laplace to transform with given conditions

Applying KCL at V(s) node:

Taking the inverse Laplace transform, we get:

i(t) = 5(1 – e^-3t/2)

Given, at t = t₀

i = 2 A

2 = 5 (1 – e^-3t/2)

t = 0.3405 sec

Concept:

A capacitor doesn’t allow a sudden change in voltage, i.e. V_c(0⁺) = V_c(0^-).

Similarly, an inductor doesn’t allow a sudden change in current, i.e. i_L(0⁺) = i_L(0^-)

Calculation:

The circuit for t = t₀^- is as shown,

V_c(t₀^-) = 5V as shown above,

At t = t₀⁺, the circuit will be as shown, in which the capacitor will be at 5V (∵ V_c(0⁺) = V_c(0^-))

To find the voltage at node ‘x’, we apply KCL to get,

Now the required current through R₁ = 5KΩ, immediately after the transition of switches is

If response due to one standard signal is known then response due to other signals can also be derived.