IEMDaily - Video Lecture Notes

Example

Identify if given circuit is inductive or capacitive

Example: When the frequency of applied emf in an LCR series circuit is less than the resonant frequency, then find the nature of the circuit.

Solution:

X_{C} = \frac{1}{ω C}

So,

X_{C} \propto \frac{1}{ω} = \frac{1}{2 π f}

X_{L} = ω L

So,

X_{L} \propto ω = 2 π f

So, as frequency decreases

X_{C}

increases So, circuit becomes capacitive circuit.

Example

Understand and find the general solution as sum of transient and steady-state solution of a RC/RL/LCR circuit

Example:
Switch

S

is closed at

t = 0

, in the circuit shown. Find the change in flux in the inductor (

L = 500 m H

) from

t = 0

to an instant when it reaches steady state.

Solution:
Before closing the switch, inductor acts as a short circuit as it is in steady state. Hence, Initial current

I_{i} = \frac{10}{10} = 1 A

Initial flux,

ϕ_{i} = L (I_{i}) = 500 m W b

= 0.5 W b

At steady state,Final current

I_{f} = \frac{20}{5} = 4 A

Final flux,

ϕ_{f} = L (I_{f}) = (0.5) \times 4 = 2 W b

∴

Δ ϕ = 1.5 W b

Example

Problems on superposition in AC circuits

Example: Find the r.m.s. value of potential due to superposition of given two alternating potentials

E_{1} = E_{0}

sin

ω t

and

E_{2} = E_{0}

cos

ω t

.

Solution:
Superposition of

E_{1}

and

E_{2} \Rightarrow E = E_{0} s i n ω t + E_{0} c o s ω t = \sqrt{2} E_{0} C o s (45^{\circ} - ω t)

R.M.S. value of

E

is given as

E_{r m s}^{2} = \frac{1}{T} \int_{0}^{T} E^{2} d t

, where

T =

\frac{2 π}{ω}

On integrating,

E_{r m s} = \sqrt{2} \frac{E_{0}}{\sqrt{2}} = E_{0}

Definition

Natural frequency for resonance in a series RLC circuits

The phenomenon of resonance is common among systems that have a tendency to oscillate at a particular frequency. This frequency is called the systems natural frequency. The resonant frequency of an RLC circuit is the frequency at which

X_{C} = X_{L}

Example

Solving problem based on resonant frequency for an RLC circuit

An LCR series circuit contains

L = 8 H

,
C

=

0.5

μ

F and R

=

100

Ω

.The resonant
frequency of the circuit is found as follows:For resonant frequency reactance should be zero
So,

X_{L} - X_{C} = 0

X_{L} = X_{C}

ω L = \frac{1}{ω C}

ω = \sqrt{\frac{1}{2 C}}

2 π f = \frac{1}{\sqrt{2 C}}

f = \frac{1}{2 π} \times \frac{1}{\sqrt{2 C}}

= \frac{1}{2 π} \times \frac{1}{\sqrt{8 \times 0.5 \times 10^{- 6}}}

= \frac{1}{2 π} \times \frac{10^{3}}{2}

= \frac{250}{π} H z

Definition

Impedance at resonance

At resonance

X_{C} = X_{L}

so impedance

Z = \sqrt{R^{2} + 0} = R

which is the minimum value of

Z

Definition

Current at resonance

At resonance, the impedance Z is minimum, then the current

i_{m} = \frac{v_{m}}{Z}

is maximum

Example

Calculating current and impedance at resonance

A series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80

μ

F, R = 40

Ω

.
Then the impedance

Z = R = 40 Ω

The peak voltage,

V_{o} = \sqrt{2} V

I_{o} = \sqrt{2} V / Z

I_{o} = 8.13 A

Definition

Applications of resonance circuits

Resonant circuits have a variety of applications, for example, in the tuning mechanism of a radio or a TV set. The antenna of a radio accepts signals from many broadcasting stations. The signals picked up in the antenna acts as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies.But to hear one particular radio station, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum.

Definition

No resonance in RL or RC circuits

It is important to note that resonance phenomenon is exhibited by a circuit only if both