Inductance, L = 3.0 H, Capacitance, C = 27 μF = 27 x 10-6 F, Resistance, R = 7.4 Ω At resonance, angular frequency of the source for the given LCR series circuit is given as ωr = 1/√(LC) = 1/√(3 x 27 x 10-6) = 111.11 rad/s Q-factor of the series: Q = ωrL/R = 111.11 x 3 /7.4 = 45.0446 To improve theRead more
Inductance, L = 3.0 H, Capacitance, C = 27 μF = 27 x 10-6 F, Resistance, R = 7.4 Ω
At resonance, angular frequency of the source for the given LCR series circuit is given as
ωr = 1/√(LC) = 1/√(3 x 27 x 10-6) = 111.11 rad/s
Q-factor of the series:
Q = ωrL/R = 111.11 x 3 /7.4 = 45.0446
To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor of 2 without changing ωr, we need to reduce R to half i.e.,
Inductance, L = 0.12 H Capacitance, C = 480 nF = 480 x 10⁻9 F Resistance, R = 23 Ω Supply voltage, V = 230 V Peak voltage is given as: Vo =√2 x 230 = 325.22 V Ans (a). Current flowing in the circuit is given by the relation, I0= V0/√[(R2 + (ωL- 1/ωC)2] Where, Io = maximum at resonance At resonance,Read more
Inductance, L = 0.12 H
Capacitance, C = 480 nF = 480 x 10⁻9 F
Resistance, R = 23 Ω
Supply voltage, V = 230 V
Peak voltage is given as: Vo =√2 x 230 = 325.22 V
Ans (a).
Current flowing in the circuit is given by the relation,
I0= V0/√[(R2 + (ωL- 1/ωC)2]
Where, Io = maximum at resonance
At resonance, we have
ωR– 1/ωRC =0
Where, ωR = Resonance angular frequency
Where , ωR = Resonance angular frequency
Therefore ωR = 1/√(LC) = 1/ √ (0.12 x 480 x 10⁻9 ) = 4166.67 rad/s
Resonant frequency , νR = ωR/2π = 4166.67 /(2 x 3.14) = 663.48 Hz
And ,maximum current, (I0)Max = V0/R = 325.22/23 = 14.14 A
Ans (b).
Maximum average power absorbed by the circuit is given as:
(Pav)Max= 1/2 x (I0)²MaxR
= 1/2 x (14.14)² x 23 = 2299.3 W
Hence, resonant frequency is 663.48 Hz.
Ans (c).
The power transferred to the circuit is half the power at resonant frequency.
Frequencies at which power transferred is half,
= ωR ± Δω
= 2π (νR ± Δν)
Where,
Δω=R/2L= 23/(2 x 0.12)= 95.83 rad/s
Hence, change in frequency,
Δν = 1/2π x Δω =95.83 /2π = 15.26 Hz
Therefore
νR +Δν = 663.48 + 15.26 = 678.74 Hz
and νR – Δν = 663.48 – 15.26 =648.22 Hz
Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half. At these frequencies, current amplitude can be given as:
I’ = 1/√ 2 x (I0)Max
= 14.14/√ 2 = 10A
Ans (d).
Q-factor of the given circuit can be obtained using the relation,
Q = ωR L/R = 4166.67 x 0.12 /23 = 21.74
Hence, the Q-factor of the given circuit is 21.74.
Average power transferred to the resistor = 788.44 W Average power transferred to the capacitor = 0 W Total power absorbed by the circuit = 788.44 W Inductance of inductor, L = 80 mH = 80 x 10-3 H Capacitance of capacitor, C = 60 μF = 60 x 10-6 F Resistance of resistor, R = 15 Ω Potential of voltageRead more
Average power transferred to the resistor = 788.44 W Average power transferred to the capacitor = 0 W Total power absorbed by the circuit = 788.44 W Inductance of inductor, L = 80 mH = 80 x 10-3 H Capacitance of capacitor, C = 60 μF = 60 x 10-6 F Resistance of resistor, R = 15 Ω Potential of voltage supply, V = 230 V Frequency of signal, ν = 50 Hz Angular frequency of signal, ω = 2πν= 2πx (50) = 100π rad/s The elements are connected in series to each other. Hence, impedance of the circuit is given as: Z = √[(R)² +(ωL -1/ωC)²]
=√[15² +((100π x 80 x 10-3 )-1/(100πx 60 x 10-6 ))²]
= √[(15)² +(25.12 -53.08)²] = 31.728 Ω
Current flowing in the circuit, I = V/Z = 230/31.728 = 7.25 A
Average power transferred to resistance is given as:
PR = I2R = (7.25)2 x 15 = 788.44 W
Average power transferred to capacitor, PC = Average power transferred to inductor, PL= 0
Total power absorbed by the circuit: = PR + PC + PL = 788.44 + 0 + 0 = 788.44 W
Hence, the total power absorbed by the circuit is 788.44 W.
Inductance, L = 80 mH = 80 x 10⁻3 H, Capacitance, C = 60 pF = 60 x 10⁻6 F Supply voltage, V = 230 V, Frequency, ν = 50 Hz, Angular frequency, ω = 2πν = 100 π rad/s Peak voltage, Vo= √2 = 230√2 V Ans (a). Maximum current is given as : Io = Vo /(1/ωL -ωC) = 230 √3 /[(100π x 80 x 10⁻3 )- 1/ (100π x 60Read more
Inductance, L = 80 mH = 80 x 10⁻3 H, Capacitance, C = 60 pF = 60 x 10⁻6 F
Supply voltage, V = 230 V,
Frequency, ν = 50 Hz,
Angular frequency, ω = 2πν = 100 π rad/s
Peak voltage, Vo= √2 = 230√2 V
Ans (a).
Maximum current is given as :
Io = Vo /(1/ωL -ωC)
= 230 √3 /[(100π x 80 x 10⁻3 )- 1/ (100π x 60 x 10⁻6 )]
= 230√2 / (8π -1000/6π) = -11.63 A
The negative sign appears because ωL <1/ωC
Amplitude of maximum current, |I0| = 11.63 A
Hence, rms value of current =I = I0/√2 =11.63 /√2 = -8.22 A
Ans (b).
Potential difference across the inductor,
VL= I x ωL
= 8.22 x 100π x 80 x 10⁻3= 206.61 V
Potential difference across the capacitor,
Vc = I x 1/ωC
= 8.22 x 1/(100π x 60 x 10⁻6 ) = 436.3 V
Ans (c).
Average power consumed by the inductor is zero as actual voltage leads the current by π/2
Ans (d).
Average power consumed by the capacitor is zero as voltage lags current by π/2
Ans (e).
The total power absorbed (averaged over one cycle) is zero.
An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where, L = 5.0 H, C = 80 μF = 80 x 10-6F, R = 40Ω Potential of the voltage source, V = 230 V IRead more
An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where,
L = 5.0 H, C = 80 μF = 80 x 10–6F, R = 40Ω
Potential of the voltage source, V = 230 V Impedance (Z) of the given parallel LCR circuit is given as:
1/Z = √(1/R² +(1/ωL -ωC)²)
Where, ω = Angular frequency Z
At resonance, (1/ωL – ωC) = 0
Therefore, ω = 1/√(LC) = 1 / √(5 x 80 x 10–6) = 50 rad/s
Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum, rms current flowing through inductor L is given as:
IL= V/ωL = 230 /(50 x 5) = 0.92 A
rms current flowing through capacitor C is given as :
IC = V/(1/ωC) =ωCV = 50 x 80 x 10–6 x 230 =0.92A
rms current flowing through resistor R is given as :
Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4 Ωμ. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘fiill width at half maximum’ by a factor of 2. Suggest a suitable way.
Inductance, L = 3.0 H, Capacitance, C = 27 μF = 27 x 10-6 F, Resistance, R = 7.4 Ω At resonance, angular frequency of the source for the given LCR series circuit is given as ωr = 1/√(LC) = 1/√(3 x 27 x 10-6) = 111.11 rad/s Q-factor of the series: Q = ωrL/R = 111.11 x 3 /7.4 = 45.0446 To improve theRead more
Inductance, L = 3.0 H,
Capacitance, C = 27 μF = 27 x 10-6 F,
Resistance, R = 7.4 Ω
At resonance, angular frequency of the source for the given LCR series circuit is given as
ωr = 1/√(LC) = 1/√(3 x 27 x 10-6) = 111.11 rad/s
Q-factor of the series:
Q = ωrL/R = 111.11 x 3 /7.4 = 45.0446
To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor of 2 without changing ωr, we need to reduce R to half i.e.,
Resistance = R/2 = 7.4/2 = 3.7 Ω
See lessA series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply. (a) What is the source frequency for which current amplitude is maximum? Obtain this maximum value. (b) What is the source frequency for which average power absorbed by the circuit is maximum? Obtain die value of this maximum power. (c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies? (d) What is the Q-factor of the given circuit?
Inductance, L = 0.12 H Capacitance, C = 480 nF = 480 x 10⁻9 F Resistance, R = 23 Ω Supply voltage, V = 230 V Peak voltage is given as: Vo =√2 x 230 = 325.22 V Ans (a). Current flowing in the circuit is given by the relation, I0= V0/√[(R2 + (ωL- 1/ωC)2] Where, Io = maximum at resonance At resonance,Read more
Inductance, L = 0.12 H
Capacitance, C = 480 nF = 480 x 10⁻9 F
Resistance, R = 23 Ω
Supply voltage, V = 230 V
Peak voltage is given as: Vo =√2 x 230 = 325.22 V
Ans (a).
Current flowing in the circuit is given by the relation,
I0= V0/√[(R2 + (ωL- 1/ωC)2]
Where, Io = maximum at resonance
At resonance, we have
ωR– 1/ωRC =0
Where, ωR = Resonance angular frequency
Where , ωR = Resonance angular frequency
Therefore ωR = 1/√(LC) = 1/ √ (0.12 x 480 x 10⁻9 ) = 4166.67 rad/s
Resonant frequency , νR = ωR/2π = 4166.67 /(2 x 3.14) = 663.48 Hz
And ,maximum current, (I0)Max = V0/R = 325.22/23 = 14.14 A
Ans (b).
Maximum average power absorbed by the circuit is given as:
(Pav)Max= 1/2 x (I0)²MaxR
= 1/2 x (14.14)² x 23 = 2299.3 W
Hence, resonant frequency is 663.48 Hz.
Ans (c).
The power transferred to the circuit is half the power at resonant frequency.
Frequencies at which power transferred is half,
= ωR ± Δω
= 2π (νR ± Δν)
Where,
Δω=R/2L= 23/(2 x 0.12)= 95.83 rad/s
Hence, change in frequency,
Δν = 1/2π x Δω =95.83 /2π = 15.26 Hz
Therefore
νR +Δν = 663.48 + 15.26 = 678.74 Hz
and νR – Δν = 663.48 – 15.26 =648.22 Hz
Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half. At these frequencies, current amplitude can be given as:
I’ = 1/√ 2 x (I0)Max
= 14.14/√ 2 = 10A
Ans (d).
Q-factor of the given circuit can be obtained using the relation,
Q = ωR L/R = 4166.67 x 0.12 /23 = 21.74
Hence, the Q-factor of the given circuit is 21.74.
See lessSuppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Average power transferred to the resistor = 788.44 W Average power transferred to the capacitor = 0 W Total power absorbed by the circuit = 788.44 W Inductance of inductor, L = 80 mH = 80 x 10-3 H Capacitance of capacitor, C = 60 μF = 60 x 10-6 F Resistance of resistor, R = 15 Ω Potential of voltageRead more
Average power transferred to the resistor = 788.44 W
Average power transferred to the capacitor = 0 W
Total power absorbed by the circuit = 788.44 W
Inductance of inductor, L = 80 mH = 80 x 10-3 H
Capacitance of capacitor, C = 60 μF = 60 x 10-6 F
Resistance of resistor, R = 15 Ω
Potential of voltage supply, V = 230 V
Frequency of signal, ν = 50 Hz
Angular frequency of signal, ω = 2 πν= 2π x (50) = 100π rad/s
The elements are connected in series to each other. Hence, impedance of the circuit is given as:
Z = √[(R)² +(ωL -1/ωC)²]
=√[15² +((100π x 80 x 10-3 )-1/(100πx 60 x 10-6 ))²]
= √[(15)² +(25.12 -53.08)²] = 31.728 Ω
Current flowing in the circuit, I = V/Z = 230/31.728 = 7.25 A
Average power transferred to resistance is given as:
PR = I2R = (7.25)2 x 15 = 788.44 W
Average power transferred to capacitor, PC = Average power transferred to inductor, PL= 0
Total power absorbed by the circuit: = PR + PC + PL = 788.44 + 0 + 0 = 788.44 W
Hence, the total power absorbed by the circuit is 788.44 W.
See lessA circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. (a) Obtain the current amplitude and mis values. (b) Obtain the rms values of potential drops across each element. (c) What is the average power transferred to the inductor? (d) What is the average power transferred to the capacitor? (e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]
Inductance, L = 80 mH = 80 x 10⁻3 H, Capacitance, C = 60 pF = 60 x 10⁻6 F Supply voltage, V = 230 V, Frequency, ν = 50 Hz, Angular frequency, ω = 2πν = 100 π rad/s Peak voltage, Vo= √2 = 230√2 V Ans (a). Maximum current is given as : Io = Vo /(1/ωL -ωC) = 230 √3 /[(100π x 80 x 10⁻3 )- 1/ (100π x 60Read more
Inductance, L = 80 mH = 80 x 10⁻3 H,
Capacitance, C = 60 pF = 60 x 10⁻6 F
Supply voltage, V = 230 V,
Frequency, ν = 50 Hz,
Angular frequency, ω = 2πν = 100 π rad/s
Peak voltage, Vo= √2 = 230√2 V
Ans (a).
Maximum current is given as :
Io = Vo /(1/ωL -ωC)
= 230 √3 /[(100π x 80 x 10⁻3 )- 1/ (100π x 60 x 10⁻6 )]
= 230√2 / (8π -1000/6π) = -11.63 A
The negative sign appears because ωL <1/ωC
Amplitude of maximum current, |I0| = 11.63 A
Hence, rms value of current =I = I0/√2 =11.63 /√2 = -8.22 A
Ans (b).
Potential difference across the inductor,
VL= I x ωL
= 8.22 x 100π x 80 x 10⁻3= 206.61 V
Potential difference across the capacitor,
Vc = I x 1/ωC
= 8.22 x 1/(100π x 60 x 10⁻6 ) = 436.3 V
Ans (c).
Average power consumed by the inductor is zero as actual voltage leads the current by π/2
Ans (d).
Average power consumed by the capacitor is zero as voltage lags current by π/2
Ans (e).
The total power absorbed (averaged over one cycle) is zero.
See lessKeeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.
An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where, L = 5.0 H, C = 80 μF = 80 x 10-6F, R = 40Ω Potential of the voltage source, V = 230 V IRead more
An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where,
L = 5.0 H, C = 80 μF = 80 x 10–6F, R = 40Ω
Potential of the voltage source, V = 230 V Impedance (Z) of the given parallel LCR circuit is given as:
1/Z = √(1/R² +(1/ωL -ωC)²)
Where, ω = Angular frequency Z
At resonance, (1/ωL – ωC) = 0
Therefore, ω = 1/√(LC) = 1 / √(5 x 80 x 10–6) = 50 rad/s
Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum, rms current flowing through inductor L is given as:
IL= V/ωL = 230 /(50 x 5) = 0.92 A
rms current flowing through capacitor C is given as :
IC = V/(1/ωC) =ωCV = 50 x 80 x 10–6 x 230 =0.92A
rms current flowing through resistor R is given as :
IR= V/R = 230/40 = 5.75A
See less