Series RL Circuit MCQ Quiz in తెలుగు - Objective Question with Answer for Series RL Circuit - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

For t = 0^-, the switch was closed and the inductor was short-circuited as shown:

\(i\left( {{0^ - }} \right) = \frac{{40}}{{{R_2}}}\)

The current flowing through the inductor for t > 0 is given as:

i(t) = 5 + 3 e^-16t

The current at t = 0⁺ will be:

i(0⁺) = 5 + 3 = 8 A

Since the inductor does not allow a sudden change in the current across it, we can write:

i (0^-) = i(0⁺) = 8 A

\(\therefore \frac{{40}}{{{R_2}}} = 8\)

R₂ = 5 Ω

For t = ∞, the switch is open and the inductor will again be short-circuited as shown:

\(i\left( \infty \right) = {i_L}\left( \infty \right) = \frac{{40}}{{{R_1} + {R_2}}}\)

Given i(t) = 5 + 3e^-16t

We can write:

\(i\left( \infty \right) = 5 = \frac{{40}}{{{R_1} + {R_2}}}\)

R₁ + R₂ = 8

With R₂ = 5 Ω

R₁ = 8 – 5 = 3 Ω

The time constant for an RL circuit is calculated as:

\(\tau = \frac{L}{{{R_{th}}}}\)

R_th = Thevenin Equivalent resistance across the inductor.

R_th = R₁ + R₂

R_th = 3 + 5 = 8 Ω

The steady-state expression for the inductor current for RL circuit is given as:

\({i_L}\left( t \right) = i\left( \infty \right) - \left[ {i\left( \infty \right) - i\left( 0 \right)} \right]{e^{ - \frac{t}{\tau }}}\)

Comparing this with the given expression, we get:

\(\tau = \frac{1}{{16}}\)

\(\therefore \frac{L}{{{R_{th}}}} = \frac{1}{{16}}\)

\(L = \frac{{{R_{th}}}}{{16}} = \frac{8}{{16}}\)

L = 0.5 H

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

Concept:

The impedance of the series RL circuit opposes the flow of alternating current.

The impedance of a series RL Circuit is nothing but the combined effect of resistance (R) and inductive reactance (XL) i.e ωL of the circuit as a whole.

The impedance Zeq in ohms is given by,

Zeq = √(R2+XL2)

Analysis:

If;

Frequency (f) of applied voltage increases → ω ↑  → XL ↑ → Zeq ↑ 

As the value of reactance increases, the phase angle also increases.

METHOD 1:

  For t<0

\(i\left( {{0^ - }} \right) = \frac{{10}}{5} = 2A\)

for t > 0 taking laplace transform

\(\begin{gathered} I\left( s \right) = \frac{{\frac{{10}}{s} + 2}}{{2 + s}}\\ \end{gathered}\)

\(I\left( s \right) = \frac{{\left( {2s + 10} \right)}}{{s\left( {s + 2} \right)}}\\ I\left( t \right) = {\mathcal{L}^{ - 1}}\left( {I\left( s \right)} \right) = 5 - 3{e^{ - 2t}}\\\)

METHOD 2:

\( i\left( t \right) = \left[ {i\left( {{0^ + }} \right) - i\left( \infty \right)} \right]{e^{\frac{{ - R}}{L}t}} + i\left( \infty \right) \\ i\left( \infty \right) = \frac{{10}}{2} = 5A,\)

\(i\left( {{0^ - }} \right) = i\left( {{0^ + }} \right) = 2\)

\( i\left( t \right) = \left( {2 - 5} \right){e^{ - \frac{{2t}}{1}}} + 5 \\ i\left( t \right) = 5 - 3{e^{ - 2t}}\\\)

For t < 0:

Initially, for t<0, the switch is closed for a long time.

The 3-ohm resistor is bypassed and the inductor is short-circuited.

Current flowing in the circuit will be:

\(I(0^-)=\frac{10}{2}=5~A\)

For t > 0:

After an infinite time when the switch is closed, the inductor is short-circuited and the current will be:

\(I(\infty)=\frac{10}{2+3}=2~A\)

The current I(t) for an inductor under transient is given by: 

\(\rm I(t)=I(\infty)+[I(0)-I(\infty)]e^{-\frac{t}{τ}}\)

τ = Time constant.

The time constant of an LR circuit is:

T= L/R

\(\rm \tau = \frac{{\frac{1}{2}}}{5} = \frac{1}{{10}}\)

\(\rm I\left( t \right) = 2 + \left[ {5 - 2} \right]{e^{ - 10t}}\)

\(\rm I\left( t \right) = 2 + 3{e^{ - 10t}}\)

At t = 0.2 sec, the current will be:

\(\rm I\left( {t = 0.2} \right) = 2 + 3{e^{ - 2}} = 2.406\ A\)

Explanation:

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).

Additional Information

Derivation for Time constant for RL Circuit:

 Consider series RL circuit,

Applying KVL to the circuit,

\(\large{V=Ri(t)+L\frac{di(t)}{dt}}\)

Taking Laplace Transformation of the equation at both sides,

\(\large{\frac{V}{s}=L[sI(s)-i(0+)]+RI(s)}\)

i(0+) = 0 (Since the current just after the switch is on, the current through the inductor will be zero.)

\(\large{\frac{V}{s}=LsI(s)+RI(s)}\)

\(\large{I(s)=\frac{V}{L} \frac{1}{s(s+R/L)}}\)

Now taking Inverse Laplace Transform to the above equation,

\(\large{i(t)=\frac{V}{R}[1-e^{\frac{-Rt}{L}}]=\frac{V}{R}[1-e^{\frac{-t}{T}}]}\)

Where, \(\large{T=\frac{L}{R}}\)

In the RL circuit, at t = T sec, the current becomes 63.2% of its final steady-state value.

At \(t = {0^ - }\)

\(i\left( {{0^ - }} \right) = 0\ A\)

Since the current across the conductor cannot change abruptly

\(i\left( {{0^ + }} \right) = 0\ A\)

Voltage across the inductor is given byA \( {V_L} = L\frac{{di}}{{dt}}\)

Applying \(KVL \to - 100 + 10\ i + 1\frac{{di}}{{dt}} = 0\)   ......(1)

\(\Rightarrow - 100 + 10\ i\left( {{0^ + }} \right) + 1\frac{{di\left( {{0^ + }} \right)}}{{dt}} = 0\)

\(\Rightarrow \frac{{di\left( {{0^ + }} \right)}}{{dt}} = 100\)

Again differentiating equation (1)

\(\Rightarrow 0 + 10\frac{{di\left( {{0^ + }} \right)}}{{dt}} + \frac{{{d^2}i\left( {{0^ + }} \right)}}{{d{t^2}}} = 0\)

\(\Rightarrow \frac{{{d^2}i\left( {{0^ + }} \right)}}{{d{t^2}}} = \frac{{ - 1000\ A}}{{{\rm{se}}{{\rm{c}}^2}}}\)

The time constant of circuit is

\(= \frac{L}{R} = \frac{{0.01}}{{10}} = 1\;msec\)

Natural response of circuit is

i_n = Ae^-1000t

For steady state:

impedance Z(jω) = R + jωL

= 10 + j10

= 10√2 ∠ 45 Ω

V_S = 120 √2 ∠ 15

\(I = \frac{{{V_s}}}{Z} = \frac{{120\sqrt 2 \angle 15}}{{10\sqrt 2 \angle 45}}\)

= 12∠-30° A

i_f = 12 sin (1000t – 30°)A

Total Response

12 sin (1000t – 30°) + Ae^-1000t

at t = 0^- i = 0

at t = 0⁺ i = 0

-12sin (-30) = Ae^°

A = 6

Total Response:

i(t) = 12 sin (1000t - 30) + 6e^-1000t

Sinsusoidal becomes zero at

1000t – 30 = 0

t = 30msec

Transient response at t = 30msec

6e^{-1000(30/1000)}

In the given circuit, the switch is on position 1 till the steady-state is reached, in the steady-state the inductor acts like a short circuit,

The circuit at t = 0^-

I(0^-) = I_L(0^-) = 50 / 10 = 5 A

As inductor does not allow a sudden change in the current so I_L (0⁺) = 5 A

Circuit at t = 0⁺

Now the circuit is a simple RL source free circuit, Inductor will start discharging and when it reaches the steady stage it will completely discharge i.e the energy stored by the inductor is zero.

So at the steady-state, the current passing through the inductor is in source free RL circuit is zero 

IL (0+) = 5 A = initial current.

I_L (∞) = 0 A = final value

The expression for I_L(t) for t ≥ 0  of source free RL circuit is given by

I_L(t) = I_L(∞) + [I_L(0⁺) - IL(∞)] e^-t/τ

Where

τ = time constant = L / R

τ = 0.2 / 10 = 0.02

I(t) = I_L(t) = 0 + ( 5 - 0) e^{-t / 0.02}

= 5 e^-50t for  t ≥ 0

Concept: 

Series RC circuit

The switch is closed at t = 0.

At t = 0⁺

At t = ∞ 

• at t = 0⁺, V_c (0⁺) = 0 V (S.C.)
• at t = ∞ , V_c (∞) = V (O.C.)

From 1st order circuit transient equation

Instantaneous current I(t) = I_c(∞) + [ I_c(0) - I_c(∞) ] e ^{- t / τ}......(1)

where  time constant, τ  = RC

Application:

\(\large I_C(0^-) = I_C(0^+) = I_c(0) ={V\over R}\)

I_c(∞) = 0

Using equation (1), 

\(\large I_c(t) = I(t) =\frac{V}{R}{e^{-t/τ}}\)

I(0) = I(0⁺) = V / R.

Hence, current though circuit, i(0⁺) = V / R.