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
The figure shows a series LCR circuit with L = 10.0 H, C = 40 μF, R = 60 Ω connected to a variable frequency 240 V source, calculate
(i) the angular frequency of the source which drives the circuit at resonance,
(ii) the current at the resonating frequency,
(iii) the rms potential drop across the inductor at resonance.
A constant current exists in an inductor-coil connected to a battery. The coil is short-circuited and the battery is removed. Show that the charge flown through the coil after the short-circuiting is the same as that which flows in one time constant before the short-circuiting.
What will be the potential difference in the circuit when direct current is passed through the circuit?
In a series, LCR circuit, obtain an expression for the resonant 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.
Assertion: When the frequency of the AC source in an LCR circuit equals the resonant frequency, the reactance of the circuit is zero, and so there is no current through the inductor or the capacitor.
Reason: The net current in the inductor and capacitor is zero.
In series LCR circuit, the phase angle between supply voltage and current is ______.
A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 kΩ, an inductor of inductive reactance XL = 250 πΩ, and an unknown capacitor. The value of capacitance to maximize the average power should be ______.
An alternating voltage of 220 V is applied across a device X. A current of 0.22 A flows in the circuit and it lags behind the applied voltage in phase by π/2 radian. When the same voltage is applied across another device Y, the current in the circuit remains the same and it is in phase with the applied voltage.
- Name the devices X and Y and,
- Calculate the current flowing in the circuit when the same voltage is applied across the series combination of X and Y.
Draw the phasor diagram for a series LRC circuit connected to an AC source.
