ATENATING UENT Important oints:. The alternating current (A) is generally expressed as ( ) I I sin ω t + φ Where i peak value of alternating current.. emf of an alternating current source is generally given by E E sin ( ω t + φ ) Where e BANω is known as peak value of alternating emf 3. Average or Mean Value: a) It is the steady current (D) which when passes through a circuit for half time period of A sends the same charge as done by the A in the same time. b) The average value of A for the complete cycle is zero. c) The average value of A for the half cycle of A is given by. I avg I.636I E imilarly, E avg is also for half cycle of A 4. M Value O Virtual Value O Effective Value: The M value of A is the steady current (D) which when flowing through a given resistance for a given time, produces the same amount of heat as produced by the A when flowing through the same resistance for the same time. I v I.77 I And Ev E.77E M value of A for half cycle is also same as above 5. A through A ure esistor E E sinωt And I I sinωt Where E and I are the peak values of Voltage and current of A respectively ince the emf and current raise or fall simultaneously they are in phase with each other.
6. A through A ure Inductor: E E sinωt And E I sin ωt ω The term ω has the units of resistance and it is called as resistance of the inductor or inductive reactance ( ) E ω ω I sin t I I sin t Where I is the peak value of current. Hence the current is lagging behind emf or emf is leading current by a phase difference of. 7. A through A ure apacitor: E E sinωt And E I cosωt / cω The term cω is called capacitive reactance where it has the dimensions of resistance. It is also called resistance of capacitor. E ω + ω + I sin t I I sin t c Where I is peak value of A Hence current leads emf or emf is lagging behind current by 8. A through ircuit: et V and V be the instantaneous voltages across the inductor and resistor respectively. V I Iω And V I Where is the inductive reactance when A flow through a pure resistor, E I + ( ω) Z ω + ( ) And tan φ ω This is the phase angle by which the emf leads the current in circuit. z is called impedance of circuit.
9. A through ircuit: I V I And V c c ω I Where capacitive reactance E V V E I I + ( ) + ω E I + ( ) ω And V Tanφ V ω This is the phase angle by which the emf lags behind the current in circuit Here Z + ω is called the impedance of circuit.. A through ircuit: a) V Iω; V I and V I ω b) E I + ω ω c) Z + ω ω is called the impedance of circuit. ω V V ω d) Tanφ This is the phase angle by which the emf leads the current. V e) If ω, then Tanφ or φ and hence emf and current are in phase. The circuit in ω this condition behaves like a pure resistor circuit. This condition is called resonance condition. f) At resonance, ω or ω ω ω n n
. Advantages of A Over D: a) The generation of A is more economic than D b) A voltages can be easily stepped up or stepped down using transformers. c) A can be transmitted to longer distances with less loss of energy. d) A can be easily converted into D by using rectifiers.. Disadvantages: a) A is more fatal and dangerous than D. b) A always flows on the outer layer of the conductor (skin effect) and hence A requires stranded wires. c) A cannot be used in electrolysis like electroplating etc. 3. Transformer: This works on the principle of mutual inductance between two circuits linked by common magnetic flux. tep up transformers converts low voltage high current into high voltage low current. tep down transformer: onverts high voltage low current into low voltage high current. E.g.: In bed lamps, we will use this type of transformer. 4. Transformer works only with ac (Alternating urrent) 5. Transformation ratio ε ε s N i s p p N p i s
Very hort Answer Questions. A transformer converts V ac into V ac. alculate the number of turns in the secondary if the primary has turns? A. V V V V n n? V V n n V n. n V. What type of transformer is used in a 6V bed lamp? A. tep down transformer. 3. What is the phenomenon involved in the working of a transformer? A. A transformer works on the principle of mutual induction. 4. What is transformer ratio? A. v v Number of turns in secondary Number of turns in primary ( N ) ( N ) 5. Write the expression for the reactance of i) an inductor and ii) a capacitor? A. Inductive reactance ω apacitive reactance ω 6. What is the phase difference between A emf and current in the following ure resistor, pure inductor and pure capacitor? A. hase difference between ac emf and current i) In pure resistor: zero ii) In pure Inductor: Voltage leads current by iii) In pure capacitor: urrent leads voltage by 9 9
7. Define power factor. On which factors does power factor depend? A. The average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle φ between them. The quantity cosφ is called power factor. p VI cosφ ower factor depends on nature of elements (esistor, Inductor, apacitor) in the circuit. 8. What is meant by wattless component of current? A. If the voltage and current differ in phase by /, then ower factor, cos φ cos 9. In this case, the current has no power. uch a current is, therefore, called wattless current. ince this current does not perform any work, this current may also be called idle current. uch a current flows only in purely inductive or in purely capacitive circuits. 9. When does a circuit have minimum impedance? A. When or ω. Then tan φ or φ tan φ or φ. Thus, there is no phase ω difference between current and potential difference. Therefore, the given circuit is equivalent to a pure resistive circuit. The impedance of such circuit is given by Z. which is minimum.. What is the phase difference between voltage and current when the power factor in series circuit is unity? A. The phase difference between voltage and current is zero. ower factor cosφ φ φ is phase difference between Voltage and current
hort Answer Questions. Obtain an expression for the current through an inductor when an A emf is applied. A. A.. through a ure Inductance: onsider a pure inductor of inductance (no resistance) connected to a source of emf ε. The instantaneous emf is given by v v sinωt m et I be the current through the circuit and di be the rate of change of current in the circuit at dt any instant. The net emf in the circuit is given by di v dt ince there is no resistance in the circuit, there is no.d. in the circuit. v m di v sinωt dt Integrating the above equation, v m di dt dt v m sinω t dt ( cosωt ) i + onstant ω ince the current is oscillatory, time independent constant does not exist. i im sin ω t The term ω has the units of resistance and it is called as inductive reactance ( ) vm i sin ωt i im sin ωt Here is the peak value of current. Hence the current is lagging behind emf or emf is leading current by a phase difference of
. Obtain an expression for the current in a capacitor when an Ac emf is applied? A. A.. through a pure capacitor: onsider a capacitor of capacity connected to a source of alternating emfε. The instantaneous emf is given by. v v sinωt m The net emf in the circuit is given by, q v (Or) q v Or q v m sinωt ( sinωt) dq d vm i ω vm cosωt dt dt vm vm (or) i cosωt Or i sin ωt i im sin ωt + + cω The term ω is called capacitive reactance and it has the dimensions of resistance. Here i is m peak value of A. Hence current leads emf or emf is lagging behind current by. 3. tate the principle on which a transformer works. Describe the working of a transformer with necessary theory? A. Transformer: A transformer converts high voltage low currents into low voltage high currents and vice-versa. Transformer works only for A. rinciple: A transformer works on the principle of mutual inductance between two coils linked by a common magnetic flux. onstruction: A transformer consists of two mutually coupled insulated coils of wire wound on a continuous iron core. One of the coils is called primary coil and the other is called secondary coil. The primary is connected to an A emf. And secondary to a load. Due to this alternating flux linkage, an emf. Is induced in the secondary due to mutual induction.
N N N N (a) (b) (c) oft iron core aminated core Working: et N and N be the number of turns in the primary and secondary coils respectively. The induced emf s produced in primary and secondary coils are given by ε N p p dφ dt and ε s N s dφ dt, ε s Ns Hence ε N p p vs Or v p N N s p Where v and v are the primary and secondary voltages. If the efficiency of the transformer is %, then i v N vsis vpip or i v N p s s s p p N N s p is called transformer ratio. If Ns > N p, then it is called a step-up transformer. If Ns < N p, then it is called a step-down transformer.
ong Answer Questions. Obtain an expression for impedance and current in series circuit. Deduce an expression for the frequency of an series resonating circuit? A. hasor Diagram olution: et an alternating emf V V sin ω t be applied to a circuit containing a resistor of resistance, a capacitor of capacitance and an inductor of inductance connected in series as shown in the fig. et i i ( ωt φ ) sin + be ac current in each element at any time where φ is the phase difference between voltage of source and the current in the circuit. et V, V, V and V represent the voltage across the inductor, resistor, capacitor and the source respectively which are shown in the phasor. The voltage equation for the circuit is V + V + V V and the amplitudes of V i, V i, V i e and V V From the diagram (B) ( ) V V + V V ( ) + ( ) i + ( ) V i i i Then i V Or i Z V ( ) + Z + impedance of the circuit where ( ) V V i i And also φ V i tan i.e. tanφ where φ is the phase angle between V and V esonance: At a resonant frequency, the total reactance of the circuit is zero and the impedance will be minimum.
From the expression Z + ( ) The impedance of the circuit Z, ( ) ω ω ω ω But ω f where f is the resonant frequency then f f At resonant frequency of series circuit, the impedance is minimum equal to and the current in the circuit is maximum.
OBEM. An ideal inductor (no internal resistance for the coil) of mh is connected in series with an A ammeter to an A source whose emf is given by sin t + / 3 V, where t is in seconds. Find the reading of the ammeter? ol: e sin t + 3 omparing with e e sin ( ω t + φ ) ω rads ; e v. 3 mh H i o eo eo x ω o o 3 e ( ) io 5 A i rms io 5 5A. The instantaneous current and instantaneous voltage across a series circuit containing resistance and inductance are given by sin( / 4) alculate the resistance? ol: i sin t A 4 omparing with i o, omparing ω rads V V o sin ω t v 4 v, ω rads o vo 4 Z Ω i o ( ) Z ω + i i sin ( ωt φ ) o v 4 sin (t) with i t A and v 4 sin (t) V. But ( ω ) + 8 ω Tanφ Tan / 4 ω ω + Ω 8 4
3. In an A circuit a condenser, a resistor and a pure inductor are connected in series across an alternator (A generator). If the voltage across them is V, 35 V and V respectively, find the voltage supplied by the alternator? ol: V v, V 35 v, V v + ( ) ( ) ( ) V V V V 35 + 35V 4. An Ac circuit contains resistance, an inductance and a capacitance connected in series across an alternator of constant voltage and variable frequency. At resonant frequency, it is found that the inductive reactance, the capacitive reactance and the resistance are equal and the current in the circuit is i. Find the current in the circuit at a frequency twice that of the resonant frequency? A. i V Z V (at resonance ) When frequency is doubled ' ' ' ' 3 Z ' + ( ) + V V V i i ' Z ' 3 3 3 5. A series resonant circuit, and. The resonant frequency is F. Another series resonant circuit contains, and. The resonant frequency is also F. If these two A. circuits are connected in series, calculate the resonant frequency? f ω and ω f ω and ω ω ω ω, ω + In series combination effective inductance () + ω And ω, ω
In series combination effective capacitance () ω esonating frequency f ' ω f c + + ω ( ) 6. In a series circuit W and the voltage and the frequency of the mains supply is V and 5 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 45. On taking out the inductor from the circuit the current leads the voltage by 45. alculate the power dissipated in the circuit? A. Ω Vrms V Z + tanφ + V rms W 7. The primary of transformer with primary to secondary turns ratio of :, is connected to an alternator of voltage V. A current of 4 A is flowing though the primary coil. Assuming that the transformer has no losses, find the secondary voltage and current are respectively? ol: N, N V. V N V V V 4V N V