Inductance, L = 2.0 H Capacitance, C = 32 μF = 32 x 10-6 F Resistance, R = 10Ω Resonant frequency is given by the relation, ωr= 1/ √ (LC) = 1/ √ (2 x 32 x 10-6 ) = 1/(8 x 10⁻³ ) = 125 rad/s Now ,Q-value of the circuit is given as : Q = 1/R√ (L/C) = (1/10) x √[2/(32 x 10-6 ) ] = 1/(10 x 4 x 10⁻³ ) =Read more
Inductance, L = 2.0 H
Capacitance, C = 32 μF = 32 x 10-6 F
Resistance, R = 10Ω
Resonant frequency is given by the relation,
ωr= 1/ √ (LC) = 1/ √ (2 x 32 x 10-6 ) = 1/(8 x 10⁻³ ) = 125 rad/s
In the inductive circuit, rms value of current, I = 15.92 A rms value of voltage, V = 220 V Hence, the net power absorbed can be obtained by the relation, P = VI cos φ Where, φ = Phase difference between V and I. For a pure inductive circuit, the phase difference between alternating voltage and currRead more
In the inductive circuit,
rms value of current, I = 15.92 A
rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,
P = VI cos φ
Where, φ = Phase difference between V and I.
For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., φ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit, rms value of current, I = 2.49 A, rms value of voltage, V = 110 V Hence, the net power absorbed can be obtained as:
P = VI Cos φ
For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e.,
Capacitance of capacitor, C = 60 μF = 60 x 10-6 F Supply voltage, V = 110 V Frequency, ν = 60 Hz Angular frequency, ω= 2πν Capacitive reactance, XC = 1/ωC = 1/2πνC =1/ (2π x 60 x 60 x 10-6) Ω rms value of current is given as : I = V/XC = 110/(1/ (2π x 60 x 60 x 10-6) ) = 110 x (2π x 60 x 60 x 10-6)=Read more
Capacitance of capacitor, C = 60 μF = 60 x 10-6 F
Supply voltage, V = 110 V
Frequency, ν = 60 Hz
Angular frequency, ω= 2πν Capacitive reactance,
XC = 1/ωC = 1/2πνC =1/ (2π x 60 x 60 x 10-6) Ω
rms value of current is given as :
I = V/XC = 110/(1/ (2π x 60 x 60 x 10-6) )
= 110 x (2π x 60 x 60 x 10-6)= 2.49 A Hence, the rms value of current is 2.49 A.
Inductance of inductor, L = 44 mH = 44 x 10⁻3 H Supply voltage, V = 220 V, Frequency, ν = 50 Hz, Angular frequency, ω = 2πν Inductive reactance, XL= ωL = 2πνL = 2π x 50 x 44 x 10-3 Ω rms value of current is given as: I = V/XL =220/(2π x 50 x 44 x 10-3) =15.92 A Hence ,the rms value of current in theRead more
Inductance of inductor, L = 44 mH = 44 x 10⁻3 H
Supply voltage, V = 220 V,
Frequency, ν = 50 Hz,
Angular frequency, ω = 2πν
Inductive reactance, XL= ωL = 2πνL = 2π x 50 x 44 x 10-3 Ω
rms value of current is given as:
I = V/XL =220/(2π x 50 x 44 x 10-3) =15.92 A
Hence ,the rms value of current in the circuit is 15.92 A
Ans (a). Peak voltage of the ac supply, Vo = 300 Vrms voltage is given as: V = V0/√ 2 = 300/√ 2 = 212.1 V Ans (b). The rms value of current is given as: I = 10 A Now, peak current is given as: I0 = √ 2 I = √ 2 x 10 = 14.1 A
Ans (a).
Peak voltage of the ac supply, Vo = 300 Vrms voltage is given as:
V = V0/√ 2 = 300/√ 2 = 212.1 V
Ans (b).
The rms value of current is given as: I = 10 A
Now, peak current is given as: I0 = √ 2 I = √ 2 x 10 = 14.1 A
Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32μF and R = 10 Ω. What is the Q-value of this circuit?
Inductance, L = 2.0 H Capacitance, C = 32 μF = 32 x 10-6 F Resistance, R = 10Ω Resonant frequency is given by the relation, ωr= 1/ √ (LC) = 1/ √ (2 x 32 x 10-6 ) = 1/(8 x 10⁻³ ) = 125 rad/s Now ,Q-value of the circuit is given as : Q = 1/R√ (L/C) = (1/10) x √[2/(32 x 10-6 ) ] = 1/(10 x 4 x 10⁻³ ) =Read more
Inductance, L = 2.0 H
Capacitance, C = 32 μF = 32 x 10-6 F
Resistance, R = 10Ω
Resonant frequency is given by the relation,
ωr= 1/ √ (LC) = 1/ √ (2 x 32 x 10-6 ) = 1/(8 x 10⁻³ ) = 125 rad/s
Now ,Q-value of the circuit is given as :
Q = 1/R√ (L/C) = (1/10) x √[2/(32 x 10-6 ) ]
= 1/(10 x 4 x 10⁻³ ) = 25
Hence, the Q-Value of this circuit is 25.
See lessIn Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
In the inductive circuit, rms value of current, I = 15.92 A rms value of voltage, V = 220 V Hence, the net power absorbed can be obtained by the relation, P = VI cos φ Where, φ = Phase difference between V and I. For a pure inductive circuit, the phase difference between alternating voltage and currRead more
In the inductive circuit,
rms value of current, I = 15.92 A
rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,
P = VI cos φ
Where, φ = Phase difference between V and I.
For a pure inductive circuit, the phase difference between alternating voltage and current is 90° i.e., φ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit, rms value of current, I = 2.49 A, rms value of voltage, V = 110 V Hence, the net power absorbed can be obtained as:
P = VI Cos φ
For a pure capacitive circuit, the phase difference between alternating voltage and current is 90° i.e.,
φ = 90°.
Hence, P = 0 i.e., the net power is zero.
See lessA 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Capacitance of capacitor, C = 60 μF = 60 x 10-6 F Supply voltage, V = 110 V Frequency, ν = 60 Hz Angular frequency, ω= 2πν Capacitive reactance, XC = 1/ωC = 1/2πνC =1/ (2π x 60 x 60 x 10-6) Ω rms value of current is given as : I = V/XC = 110/(1/ (2π x 60 x 60 x 10-6) ) = 110 x (2π x 60 x 60 x 10-6)=Read more
Capacitance of capacitor, C = 60 μF = 60 x 10-6 F
Supply voltage, V = 110 V
Frequency, ν = 60 Hz
Angular frequency, ω= 2πν
Capacitive reactance,
XC = 1/ωC = 1/2πνC =1/ (2π x 60 x 60 x 10-6) Ω
rms value of current is given as :
I = V/XC = 110/(1/ (2π x 60 x 60 x 10-6) )
= 110 x (2π x 60 x 60 x 10-6)= 2.49 A
See lessHence, the rms value of current is 2.49 A.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Inductance of inductor, L = 44 mH = 44 x 10⁻3 H Supply voltage, V = 220 V, Frequency, ν = 50 Hz, Angular frequency, ω = 2πν Inductive reactance, XL= ωL = 2πνL = 2π x 50 x 44 x 10-3 Ω rms value of current is given as: I = V/XL =220/(2π x 50 x 44 x 10-3) =15.92 A Hence ,the rms value of current in theRead more
Inductance of inductor, L = 44 mH = 44 x 10⁻3 H
Supply voltage, V = 220 V,
Frequency, ν = 50 Hz,
Angular frequency, ω = 2πν
Inductive reactance, XL= ωL = 2πνL = 2π x 50 x 44 x 10-3 Ω
rms value of current is given as:
I = V/XL =220/(2π x 50 x 44 x 10-3) =15.92 A
Hence ,the rms value of current in the circuit is 15.92 A
See less(a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Ans (a). Peak voltage of the ac supply, Vo = 300 Vrms voltage is given as: V = V0/√ 2 = 300/√ 2 = 212.1 V Ans (b). The rms value of current is given as: I = 10 A Now, peak current is given as: I0 = √ 2 I = √ 2 x 10 = 14.1 A
Ans (a).
Peak voltage of the ac supply, Vo = 300 Vrms voltage is given as:
V = V0/√ 2 = 300/√ 2 = 212.1 V
Ans (b).
The rms value of current is given as: I = 10 A
Now, peak current is given as: I0 = √ 2 I = √ 2 x 10 = 14.1 A
See less