Electric Circuit Analysis: Unit V: Resonance and coupled circuits

Worked examples on series resonance

Electric Circuit Analysis: Unit V: Resonance and coupled circuits : Worked examples On series resonance

WORKED EXAMPLES

ON SERIES RESONANCE

Example 1 A series RLC circuit with R = 102, L = 1 mH and C = 1000 picofarads is connected across a sinusoidal source of 20V with variable frequency (a) Compute the resonant frequency of the circuit (b) Find the Q factor of the circuit at the resonant frequency, (c) Determine the half power frequencies.

Solution: Given R = 10 Ω, L= 1 mH, C = 1000 pF, E = 20 V


 

Example 2 A series RLC circuit with Q = 250 is resonant at 1.5 MHZ. Find the frequencies at half power points and also band width.

Solution:


 

Example 3 A coil of resistance 2 Ω and inductance 0.01 H is connected in series with a capacitor C. If maximum current occurs at 25 Hz find C.

Solution: When the current is maximum the circuit is at resonance. Therefore, the given frequency is resonance frequency.


 

Example 4 What is the resonant frequency and Bandwidth of a series RLC circuit given R = 552, L=40 mH, C=1 μF.

Solution:


 

Example 5 A coil of resistance 402 and inductance 0.75 H forms part of a series circuit for which the resonant frequency is 55 c/s. If the supply is 250V, 50 c/s find (i) the line current, (ii) the p.f., (iii) voltage across the coil.

Solution:


 

Example 6 A constant voltage at a frequency of 1 MHZ is applied to an inductor in series with a variable capacitor. When the capacitor is set to 500 pF, the current has the maximum value, while it is reduced to one half when the capacitance is 600 pF. Find (i) the resistance, (ii) the inductance, (iii) the Q factor of the inductor.

Solution:


Let R be the resistance and E be the applied voltage.

At resonance, current I = E/R

Given that, When C = 600 pF, the current is reduced to half of current at resonance. Let Z be the impedance when the current is half at resonance. Therefore,



Example 7 A series RLC circuit has R = 2002, L = 20mII and C = 0.5uF. Calculate the maximum current in the circuit and voltage across L and C for an input voltage 200 0°V.

Solution: Since the current is maximum, the circuit is at resonance.


Note: |EL| = | Ec| at resonance. But they are always in anti-phase.

 

Example 8 A coil of inductance 9H and resistance 502 in series with a condenser is supplied at constant voltage from variable frequency source. If the maximum current is 1 amp at 75 c/s, find the frequency when the current is 0.5 amps.

Solution: Given: R = 50 Ω

L = 9H

Maximum current = 1 amp

F0 = 75 Hz

It is required to find the frequency, when the current is 0.5 A.

Maximum current = I = E / R

E = IR = 1 × 50 = 50 volts

When the current is 0.5A, let the impedance be Z.


 [Note: The student is advised not to mistake the f1 and f2 as half power frequencies. Only for convenience we have taken the symbols f1 and f2 for frequencies.]

 

6. Frequency at which Maximum Values of Ec and EL Occur

Case (a): To find the frequency at which Ec is maximum.

The voltage across the capacitor = | Ec|


We have to find out the frequency at which Ec is maximum i.e., at that frequency

dEc / d ω = 0.

Before differentiating the above equation, it is desirable to eliminate the radical sign from the denominator. This can be done by differentiating E2 with respect to o, since when E2 maximum Ec

will also be maximum.

By squaring equation (xvi), we get


At this frequency of o (in rad/sec) Ec is maximum. This ω is lesser than wc. From the above equation,

we say that at  EC maximum

Case (b): To find the frequency at which EL is maximum.

The voltage across L = |EL| = |1| |XL|


At this frequency ω , ELis maximum. This frequency is more than ωo.

 

Example 9 A series circuit with R = 50 2; L = 0.05H and C = 20 micro farads has an applied voltage of 100 0° with a variable frequency. Find the maximum voltage across the inductor as the frequency is varied.

Solution: Given : R 50Ω L = 0.05H

C = 20 × 10-6 F; E = 100 ∠0o volts.

Frequency is variable.

The frequency at which EL is maximum is given by the formula



Electric Circuit Analysis: Unit V: Resonance and coupled circuits : Tag: : - Worked examples on series resonance