Advertisements
Advertisements
प्रश्न
Using the phasor diagram, derive the expression for the current flowing in an ideal inductor connected to an a.c. source of voltage, v= vo sin ωt. Hence plot graphs showing the variation of (i) applied voltage and (ii) the current as a function of ωt.
Advertisements
उत्तर
Figure (a) shows an ac source connected to an inductor. Usually, inductors have appreciable resistance in their windings, but we shall assume that this inductor has negligible resistance. Thus, the circuit is a purely inductive ac circuit. Let the voltage across the source be V = Vm sin ωt. Using the Kirchhoff’s loop rule, Σε (t) = 0, and since there is no resistor in the circuit,
`v - L(di)/(dt) = 0` ...(i)
where the second term is the self-induced Faraday emf in the inductor; and L is the self-inductance of the inductor. The negative sign follows from Lenz’s law.
From equation (i) we have
`(di)/(dt) = v/L = v_m/L sin ωt` ...(ii)
Equation (ii) implies that the equation for i(t), the current as a function of time, must be such that its slope di/dt is a sinusoidally varying quantity, with the same phase as the source voltage and an amplitude given by .vm/L
To obtain the current, we integrate di/dt with respect to time:
`int_ (d"i")/(d"t") d"t" = "v"_"m"/"L" int_ sin(wt)dt`
and get
`"i" = -("v"_"m")/(w"L") cos (wt) + cons tan t`
The integration constant has the dimension of current and is time-independent. Since the source has an emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero.
Using
`-cos (wt) = sin(wt - pi/(2)), "we have"`
`"i" - "i"_"m" sin(wt - pi/(2))`
Where `"i"_"m" = ("v"_"m")/(w"L")` is the amplitude of the current.
The quantity ωL is analogous to the resistance and is called inductive reactance, denoted by XL :
`"X"_"L" = w"L"`
The amplitude of the current is, then
`"i"_"m" = ("v"_"m")/("X"_"L")`
The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm (Ω). The inductive reactance limits the current in a purely inductive circuit in the same way as the resistance limits the current in a purely resistive circuit. The inductive reactance is directly proportional to the inductance and to the frequency of the current.
Fig. (b) A Phasor diagram for the circuit in Fig.(a) Fog. (c) Graph of v and i versus wt.
APPEARS IN
संबंधित प्रश्न
An L-R circuit has L = 1.0 H and R = 20 Ω. It is connected across an emf of 2.0 V at t = 0. Find di/dt at (a) t = 100 ms, (b) t = 200 ms and (c) t = 1.0 s.
An LR circuit with emf ε is connected at t = 0. (a) Find the charge Q which flows through the battery during 0 to t. (b) Calculate the work done by the battery during this period. (c) Find the heat developed during this period. (d) Find the magnetic field energy stored in the circuit at time t. (e) Verify that the results in the three parts above are consistent with energy conservation.
The current in a discharging LR circuit without the battery drops from 2.0 A to 1.0 A in 0.10 s. (a) Find the time constant of the circuit. (b) If the inductance of the circuit 4.0 H, what is its resistance?
(i) An a.c. source of emf ε = 200 sin omegat is connected to a resistor of 50 Ω . calculate :
(1) Average current (`"I"_("avg")`)
(2) Root mean square (rms) value of emf
(ii) State any two characteristics of resonance in an LCR series circuit.
Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 µF, R = 40 Ω.
- Determine the source frequency which drives the circuit in resonance.
- Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
- Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Keeping 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 for this frequency.
For a series LCR-circuit, the power loss at resonance is ______.
To reduce the resonant frequency in an LCR series circuit with a generator
In an LCR circuit having L = 8 henery. C = 0.5 µF and R = 100 ohm in series, the resonance frequency in radian/sec is
A series LCR circuit containing a 5.0 H inductor, 80 µF capacitors, and 40 Ω resistor is connected to a 230 V variable frequency ac source. The angular frequencies of the source at which power is transferred to the circuit are half the power at the resonant angular frequency are likely to be ______.
