CLUSTER LEVEL WORK SHOP

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Coral Holt
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1 CLUSTER LEVEL WORK SHOP SUBJECT PHYSICS QUESTION BANK (ALTERNATING CURRENT ) DATE: 0/08/06 What is the phase difference between the voltage across the inductance and capacitor in series AC circuit? Ans. Phase difference 80 0 In a series of LCR circuit V L = V C V R, What is the value of power factor for this circuit? power factor = 3 The instantaneous current and voltage of an a.c. circuit are given by i = 0 sin 300 t A and V = 00 sin 300 t V. What is the power dissipation in the circuit? It is given that: i = 0 sin 300 t A V = 00 sin 300 t V i 0 = 0 A and V 0 = 00 V Average power dissipation = V 0 i 0 = 00 0 = 000 W 4 In a series LCR circuit the voltage across an inductor, a capacitor and a resistor are 0V, 0V and 60V respectively. What is the phase difference between the applied voltage and the current in the circuit? VL= VC,(XL = XC) than circuit is said to be in resonance Z = R,Therefore phase difference V And I is What is the average value of the emf for the shaded part of graph? average value of the emf for the shaded part = V 0 = = 00V 6 The instantaneous current from an a.c. source is I = 5 sin (34 t) ampere. What are the average and rms values of the current? ANS: average value of current = 0, rms value of current I 0 / = 5 A = 3.53 A 7 What is the inductive reactance of a coil if current through it is 800 ma and the voltage across it is 40 V? ANS: X L = V/I = 40/ 800 m A = 40000/800 = 50 ohm 8 A transformer has 300 primary turns and 400 secondary turns.if the primary supply voltage is 30 V, what is the secondary voltage? N P /N S = V P / V S ; = /300 =.84 kv

9 A.C. power is transmitted from one station to another at highest possible voltage. Why? ANS: As the voltage of A.C. power transmitted from one station to another is very large, the magnitude of the current is very low.

2 9 A.C. power is transmitted from one station to another at highest possible voltage. Why? ANS: As the voltage of A.C. power transmitted from one station to another is very large, the magnitude of the current is very low. Therefore, the power loss is reduced in transmission of power as the power loss is given by P = I R 0 What is wattless current? In the purely inductive and capacitive circuit power loss is zero in such a circuit current fallowing is called wattles current Why is the use of A.C. voltage preferred over D.C.voltage? Give two reasons. Solution: The use of A.C. voltage is preferred over the use of D.C. voltage because of the following reasons: i) The loss energy in transmitting the A.C. voltage over long distances with the help of step up transformers is negligible as compared to D.C. voltage. ii) A.C. voltage can be stepped up and stepped down as per the requirement using a transformer. What is average value of AC voltage represented by V = 0 sin34t volts over the time interval a. 0 to π/ω b. 0 to π/ω we know that T = π/ω Here given t = π/ω = T/ i.e. half period, V av = V o / π = 7/ V o = = 40 V Over a complete cycle, V av = 0. 3 Calculate power factor of an AC circuits having a resistance R and Inductance L connected in series. Given angular frequency of source is ω. Z = R + X L = R + ω L Power factor = cos(ᶲ) = R z = R R +ω L 4 In the figure below if the frequency of the A.C mains is increased. How will the current be affected? R A.C Mains R is not affected by frequency. Hence no change in current. 5 A bulb B and a capacitor C are connected in series with a.c. mains. The bulb glows with some brightness. How will the glow of the bulb will change when a mica sheet is introduced between the plates of capacitor. C B AC A By introducing mica sheet between the plates of capacitor, the capacitance increases. Hence Capacitive reactance X C =/ ω C,decreases. Glow of bulb decreases

3 6 A bulb B and an inductor L are connected in series with a.c mains through key. The switch is closed and after some time an iron rod is inserted into the interior of the inductor. How will the glow of the bulb will change? L B AC A ANS: XL = ωl, Inductance increases. Hence inductive reactance increases. Low amount of voltage appears across the bulb. Hence glow of bulb decreases. 7 Prove that an ideal capacitor in an AC does not dissipated power. P = Vrms I rms cos, for ideal capacitor R = 0, therefore cos = 0, P = 0 8 In a series LCR circuit the value of inductance is kept fixed at resonance but the resistance is doubled. How will it affect the sharpness of resonance? Q = ω r L/ R; Q = ω r L/ R = Q = ω r L/ R Q = Q/, half 9 0 An electric bulb B and a parallel plate capacitor C are connected in series to the a.c.mains as 3 shown in the given figure. The bulb glows with some brightness. How will the glow of the bulb be affected on introducing a dielectric slab between the plates of the capacitor? Give reasons in support of your answer. Solution: The bulb will glow brighter. Reasons: The impedance of a capacitor is, X C = without dielectric. If a dielectric is introduced inside a capacitor, then the new capacitance will be C = KC and the new impedance will be X C =. Therefore, the impedance has decreased. This will result in higher current through the circuit and the bulb will glow brighter.

4 In a series LCR circuit, define the quality factor (Q) at resonance. Illustrate its significance by giving one example. Show that power dissipated at resonance in LCR circuit is the maximum. 3 Solution: The quality factor (Q) is defined as: Where, Resonant frequency L Inductance R Resistance Another expression for quality factor is Where, is the band width. So, larger the value of Q, the smaller the band width and sharper is the resonance. Power = Hence, power = Power dissipated will be the maximum at resonance 3 A resistor of 00 Ω and a capacitor of 40 μf are connected in series to 0 V a.c. source with angular frequency (ω) = 300 Hz. Calculate the voltages (rms) across the resistor and the capacitor. Why is the algebraic sum of these voltages more than the source voltage? How do you resolve this paradox? 3

5 Voltage drop across resistor = IR =.0 00 = 04 V Voltage drop across capacitor = IX C Total voltage drop = = 89 V Kirchoff s voltage law will be valid only if we consider instantaneous values of voltage. 85 V and 04 V are rms values of voltage and hence, when they are added there value becomes more than the source voltage 4 A light bulb is rated 00 W for 0 V ac supply of 50 Hz. Calculate (i) The resistance of the bulb; (ii) The rms current through the bulb P = 00 W V rms = 0 V (a) Resistance, R 3 (b) P = I rms V rms 5 An alternative voltage given by V = 40 sin 34t is connected across a pure resistor of 50 Ω. Find (i) The frequency of the source. (ii) The rms current through the resistor. Given V = 40 sin 34t, R = 50Ω (i) Comparing with V = V 0 sin t Thus, V 0 = 40 V ω = 34 π = A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation. 3 Solution: Let an alternating Emf is E = E 0 sin t is applied to a series combination of inductor L, capacitor C and resistance R. Since all three of them are connected in series the current through them is same. But the voltage across each element has a different phase relation

6 with current. The potential difference V L, V C and V R across L, C and R at any instant is given by V L = IX L, V C = IX C and V R = I R Where I is the current at that instant. X L is inductive reactance and X C is capacitive reactance. V R is in phase with I. V L leads I by 90 and V C lags behind I by 90 In the phases diagram, V L and V C are opposite to each other. If V L > V C then resultant (V L V C ) is represent by OD. OR represent the resultant of V R and (V L V C ). It is equal to the applied Emf E. The term is called impedance Z of the LCR circuit.

7 Emf leads current by a phase angle When resonance takes place Impedance of circuit becomes equal to R. Current becomes maximum and is equal to This is the condition for resonance. When at resonance f = f 0 the current in the circuit is maximum and hence impedance of the circuit is maximum for values of f less than or greater than f 0 comparatively small current flames in the circuit. 7 a) For a given a.c., i = i m sin t, show that the average power dissipated in a resistor R over a 3 complete cycle is R. (b) A light bulb is rated at 00 W for a 0 V a.c. supply. Calculate the resistance of the bulb. Solution: (a)

8 (b) Power of the bulb, P = 00 W and Voltage, V = 0 V a.c source generating a voltage is connected to a capacitor of An 5 capacitance C. Find the expression for the current i, flowing through it, plot a graph of v and i versus ωt to show that the current is, ahead of the voltage. A resistor of 00 Ω and a capacitor of 5 μf are connected in series to a 0 V, 50 Hz a.c source. Calculate the current in the circuit and the rms voltage across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. A.C source containing capacitor: Alternating emf supplied is: Potential difference across the plates of capacitor At every instant, the potential difference V must be equal to the emf applied i.e., Or, q = C It I is instantaneous value of current in the circuit at instant t, then

Numerical: Here, r = 00 Ω, C = 5 μf = 5 0 6 F E = 0 V, f = 50 Hz, I =? This is because these voltages are not in same phase and they cannot be added like ordinary numbers.

9 Numerical: Here, r = 00 Ω, C = 5 μf = F E = 0 V, f = 50 Hz, I =? This is because these voltages are not in same phase and they cannot be added like ordinary numbers. 3 Explain briefly, with the help of a labelled diagram, the basic principle of the working of an a.c generator. In an a.c generator, coil of N turns and area A is rotated at V revolutions per second in a uniform magnetic field B. Write the expression for the emf produced. A 00-turn coil of area 0. m rotates at half a revolution per second. It is placed in a magnetic field 0.0 T perpendicular to the axis of rotation of the coil. Calculate the maximum voltage generated in the coil. A.C generator principle: Whenever a closed coil is rotated in a uniform magnetic field about an axis perpendicular to the field, the magnetic flux linked with coil changes and an induced emf is set up across its ends. 5 The construction of an ac generator is shown in the figure. Initially, the coil ABCD is horizontal. The coil starts rotating clockwise and the arm AB moves up whilecd moves down. By Fleming s right hand rule, the induced current flows along ABCD. In second half rotation, the arm CD moves up and AB moves down. The induced current

10 flows in the opposite direction that is along DCBA. Thus, an alternating current flows in the circuit. The magnetic flux linked with the coil at any instant is Φ = NBA cos ωt Induced emf will be Or E = E 0 sin ωt, where E 0 = NBA ω = Peak value of induced emf Numerical: N = 00, A = 0. m, B = 0.0T Maximum voltage, 33 (a) Derive an expression for the average power consumed in a series LCR circuit connected to a.c. source in which the phase difference between the voltage and the current in the circuit is Φ. (b) Define the quality factor in an a.c. circuit. Why should the quality factor have high value in receiving circuits? Name the factors on which it depends. (a) Power in ac circuit Voltage v in an ac circuit is: 5 which drives through the circuit a current i i = i m sin (ωt +Φ), where and Power Calculating the average power, it is observed that the average of the term cos (ωt + Φ) is equal to zero. Thus, Average power,

11 (b) The ratio is called the quality factor or Q-factor. The quality factor has high value in receiving circuits in order to get a sharp gain for the desired channel frequency. The quality factor depends on the values of the following: i. Inductance ii. Resistance iii. Capacitance 34 (a) Derive the relationship between the peak and the rms value of current in an a.c. circuit. (b) Describe briefly, with the help of labelled diagram, working of a step-up transformer. A step-up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain. OR (a) The instantaneous power dissipated in the resistor is The average value of p over a cycle is: 5 are constants. Therefore, By trigonometric identity, The average value of cos ωt is zero. We have: Thus, The rms value in the ac power is expressed in the same form as dc power root mean square or effective current and is denoted by I rms. Peak current is Therefore,

12 (b) In a transformer with N s secondary turns and N p primary turns, induced emf or voltage E s is: Back emf = E p = E P = V P E s = V s

13 Thus, V s = (i) Dividing equations (i) and (ii), we obtain If the transformer is 00% efficient, then Thus, combining the above equations, If N s > N p, then the transformer is said to be step-up transformer because the voltage is stepped up in the secondary coil. No, the transformer does not violate the principal of conservation of energies. This can be easily observed by the following equation: Power consumed in both the coils is the same as even if the voltage increases or current increases, their product at any instant remains the same. 35 A series LCR circuit is connected to a source having voltage v = v m sin ωt. Derive the expression for the instantaneous current I and its phase relationship to the applied voltage. Obtain the condition for resonance to occur. Define power factor. State the conditions under which it is (i) maximum and (ii) minimum. 5 v = v m sin ωt Let the current in the circuit be led the applied voltage by an angleφ. The Kirchhoff s voltage law gives. It is given that v = v m sin ωt (applied voltage) On solving the equation, we obtain On substituting these values in equation (), we obtain

14 Let This gives and On substituting this in equation (), we obtain On comparing the two sides, we obtain Or And The condition for resonance to occur For resonance to occur, the value of i m has to be the maximum. The value of i m will be the maximum when

15 Power factor = cos Φ Where, (i) Conditions for maximum power factor (i.e., cos Φ = ) i. X C = X L Or ii. R = 0 (ii) Conditions for minimum power factor iii. When the circuit is purely inductive iv. When the circuit is purely capacitive

Alternating Current Circuits

Alternating Current Circuits AC Circuit An AC circuit consists of a combination of circuit elements and an AC generator or source. The output of an AC generator is sinusoidal and varies with time according