Test: Step Response of First Order Circuits - 1 - Electrical Engineering (EE) MCQ

Concept:

The admittance triangle is also represented similarly to the impedance triangle. As the impedance (Z) of the circuit has two rectangular components, resistance (R) and reactance (X).

Similarly, admittance (Y) also has two components, conductance (G) and susceptance (B).

Then Power factor will be:

PF = cosθ  = G/Y

Calculation:

Given;

G = 30 siemens

B = 40 Siemens

Y = √(30² + 40²) = 50 Siemens

PF = cosθ = 30/50 = 0.6

The expression of current in a series RL circuit is:

V = 60 V, R = 30, L = 15 H

Current through the inductor is,

⇒ I = 2(1 – e^-2t)

An L-R Series Circuit consists basically of an inductor of inductance L, connected in series with a resistor of resistance R. The resistance “R” is the resistive value of the wire turns or loops that go into making up the inductors coil.

Time constant (τ) –

The time constant is defined as the time required for the circuit to reach 63.2% of the final value (steady-state value).

Given circuit diagram:

The above circuit can be redrawn as, (the above circuit forms a balanced bridge condition)

The equivalent resistance of all the 5 Ω resistors is 5 Ω 

Current in the circuit will be 

i(t) = I e^-4τ

Time constant τ = RC = 5 secs

Energy transferred from the DC source,

Concept:

The Series R-L circuit with DC source is shown below as.

Formula:

The  voltage across the resistor is 

V_R = i.R

The voltage across the Inductor is 

The formula for alternating current passes through the R-L circuit is

Calculation:

The phasor diagram is drawn as:

1) There is no phase difference between the applied voltage and the voltage across R and C in parallel.

2) The current through the resistive branch is in phase with the applied signal.

3) But the current through the capacitive branch leads its voltage Vc by 90 degrees.

.

Concept:-  A coil is made up of inductive reactance and Resistance.

i.e. When it is connected to the DC supply, the inductor will act as a short circuit

Hence, Resistance of the coil =  Ω

For AC supply, the inductor will offer finite reactance that will depend on the frequency of the supply.

i.e, X_L = ωL

Now,

Z = V/I = 32/5 = 6.4 Ω

We know that,

Power Factor = R/Z = 5/6.4 = 0.78

And,

X² = Z² - R²

Hence,

Concept:

For RL load, θ = tan^-1 (ωL/R) and,

The condition for transient free response is, θ - α = 0 ⇒ θ = α

Calculation:

V = 100 sin (200t + α)

R = 20Ω, L = 0.1 H

⇒ α = 45°

Concept:

For a series RL circuit, the net impedance is given by:

Z = R + j (X_L)

X_L = Inductive Reactance given by:

X_L = ωL

The magnitude of the impedance is:

Calculation:

With R = 3 Ω and XL = 4 Ω, we get:

|Z| = 5 Ω

Concept:

A capacitor doesn’t allow a sudden change in voltage across it, i.e.

V_c(0^-) = V_c(0⁺)

Calculation:

Given circuit can be drawn as:

⇒ V_c(0^-) = 0V (because the switch is open)

At t = 0⁺ (i.e. at the instant when the switch is closed)

V_c(0⁺) = 0 V