Alternating Current -03-SUBJECTIVE SOLVED

SOLVED SUBJECTIVE PROBLEMS Problem 1. A resistance R and inductance L and a capacitor C all are connected in series with an AC supply. The resistance of R is 16 ohm and for a given frequency, the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm. If the current in the circuit is 5 amp., find (a) The potential difference across R, L and C (b) The impedance of the circuit (c) The voltage of AC supply (d) Phase angle. Solution: (a) Potential difference across the resistance VR  i R  516  80 Volt Potential difference across the inductor VL  i (L)  5 24  120 Volt Potential difference across the capacitor VC  i (l / C)  512  60 Volt (b) Z  = [(16)2  (24 12)2 ]  20 ohm (c) The voltage of AC supply is given by E = I Z = 5 × 20 = 100 volt (d) Phase angle   tan1 L  (1/ C)   tan1  24 12   R   16  = tan1(0.75)  36º 46 ' . Problem 2. A circuit draws a power of 550 watt from a source of 220 volt, 50 Hz. The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. To make the power factor of the circuit as 1.0, what capacitance will have to be connected with it. Solution: As the current lags behind the potential difference, the circuit contains resistance and induc- tance. Power, P = vrms × irms × cos  Here, irms  Vrms , where Z  Z  V 2  cos  v2  cos   P rms or Z  rms Z P (220)2  0.8 So, Z   70.4 ohm 550 Now, power factor cos   R Z or R = Z cos   R = 70.4 × 0.8 = 56.32 ohm Further, Z2  R 2  (L)2 or (L)  Or L   42.2 ohm When the capacitor is connected in the circuit, Z  and cos   R when cos   1, L  1 C  C  1  1 (L) 2f (L)  (2 3.14  50)  (42.2)  75106 F  75F . Problem 3. Aseries L-C-R circuit is connected to an AC source of 220 V and 50 Hz. as shown in figure. If the readings of the three voltmeters V1, V2 and V3 are 65 V, 415 V and 204 V respectively, calculate, (i) The current in the circuit (ii) The value of inductor (iii) The value of the capacitor C and 220V 50 Hz (iv) The value of C (for the same L) required to produce resonance. Solution: (i) Here VR = irms R, where irms is the rms value of current in the circuit.  i  VR  65  0.65 amp (ii) rms V  i R 100  X or X  VL L rms L L irms  XL  204 0.65  313.85 Now, XL  L  2fL or L  XL 2f L  313.85  1.0 henry 2 50 X  VC  415  638.46 (iii) irms 0.65 X  1  1 C C 2fC  C  1 2 50  638.46  5106  5F (iv) Let ‘C’ be the capacitance of capacitor that will produce resonance with inductor L = 1.0 henry. Then f  or C'  1 42f 2L  1 42 502 1.0  10.1106 F  10.1F . Problem 4. A 750 Hz., 20 V source is connected to a resistance of 100 ohm, an inductance of 0.1803 henry and a capacitance of 10 microfarad all in series. Calculate the time in which the resistance (thermal capacity 2 J/ºC) will get heated by 10ºC. Solution: As in this problem, XL  L  2fL  2 750  0.1803  849.2 and, X  1  1  1  21.2 C C 2fC 2 750 105 so, X  XL  XC  849.2  21.2  828 and hence, Z    834 but as in case of ac, P  V I cos   V  Vrms  R av rms rms rms Z Z  V 2  20 2 i.e., Pav  rms Z  R   834  100  0.0575W     and as, U  P  t  mc  (TC) t  (TC)    2 10  348s □ 5.8 min. P 0.0575 Problem 5. An LCR series circuit with 100  resistance is connected to an ac source of 200 V and angular frequency 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60º. When only the inductance is removed, the current leads the voltage by 60º. Calculate the current and the power dissipated in the LCR circuit. Solution: When capacitance is removed, the circuit becomes L-R with, tan   XL R i.e., XL  R tan   100  and when inductance is removed the circuit becomes C-R with, tan   XC R i.e., XC  R tan   100  as here XL  XC so the circuit is in series resonance and hence as X = XL  XC = 0, i.e., Z  So, Irms  R ,  Vrms  Vrms  200  2 A Z R 100 And, Pav  Vrms Irms cos   200  2 1  400W Problem 6. An inductor of 20103 henry, a capacitor 100 F and a resistor 50  are connected in series across a source of emf V = 10 sin (314t). Find the energy dissipated in the circuit in 20 minutes. If resistance is removed from the circuit and the value of inductance is doubled, then find the variation of current with time in the new circuit. Solution: Here, time of 1 cycle T = 2   2 314 =1/50 s. So, we have to calculate the average energy as time >> T. Energy consumed in time t = (Vrms Irms cos)t  I0 . V0 . R t Z L R C V2R E  0 t I  V0  2Z2  0 Z    Now, Z  10 sin 314t    56 ohm  102  50  20  60  Energy consumed When resistance is removed, 2  3153.6 joule cos  R Z'  0 or    / 2 Z  1  L '  1  314  40 103 c 314 104 = 19.3  I  E0 sin(t  ) Z'  10 sin(314t   / 2) 19.3 = 0.52 cos 314 t Amp. Problem 7. A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of source is constant at 10 V. Box P contains a capacitance of 1 F in series with a resistance of 32  . Coil Q has a self-inductance 4.9 mH and a resistance 68  in series. The frequency is adjusted so that the maximum current flows in P and Q. Find the imped- ance of P and Q at this frequency. Also find the voltage across P and Q respectively. Solution: As this circuit is a series LCR circuit, current will be maximum at resonance, i.e., 105 rad     7 s E 10V 1 with Imax  0   A R (32 68) 10 So the impedance for box P   1 2 1/ 2   7 2 1/ 2 Z  R 2    (32)2     77 P 1  C   105 106          and for coil Q and hence Z [R2 (L)2 ]1/ 2 [(68)2 (4.9103 105 / 7)2 ]1/ 2   97.6 V  IZ  1  (77)  7.7V (potential drop across P) P P 10 and V  IZ  1  (97.6)  9.76V (potential drop across Q) Q Q 10 Problem 8. An LCR circuit has L = 10 mH, R = 3  and C = 1 F connected in series to a source of 15 cos t V. Calculate the current amplitude and the average power dissipated per cycle at a frequency that is 10% lower than the resonance frequency. Solution: As here resonance frequency,     104 rad so, 0     10   s 9   9103 rad 0 100 0 10 0 s and hence, X  L  9 103 102  90 X  1  1  111.11 C C 9 103 106 so, X  XC  XL  111.11 90  21.11 and hence, Z   i.e., Z   21.32 and as here E = 15 cos t , i.e., E0 =15 V, I  E0  15  0.704 A 0 Z 21.32 The average power dissipated, P  V I cos  (I  Z)  I  R av rms rms rms rms Z i.e., P  I2 R  1 I2R as I  I0  av rms 2 0  rms 2  so, Pav  1 (0.704)2  3  0.74 W 2 Now as f    9 103 cycle 2 2 s So, Pav  0.74 J / s  2 0.74 J cycle (9 103 / 2) cycle / s 9103 cycle i.e., Pav  5.16104 J . cycle cycle Problem 9. A series LCR circuit containing a resistance of 120  has angular resonance frequency 4105 rad s1 . At resonance the voltages across resistance and inductance are 60 V and 40 V respectively. Find the values of L and C. At what frequency the current in the circuit lags the voltage by 45º ? Solution: At resonance as X = 0, I  V  60  1 A and as V  IX R 120 2  IL L  VL so, L L L  40 I  0.2 mH (1/ 2)  4105 and as   1 , C  1 L2 i.e., C  1 0.2 103  (4 105 )2  1 F 32 Now in case of series LCR circuit, tan   XL  XC R so current will lag the applied voltage by 45º if, L  1 i.e., tan 45  C R 1120   2 104  1 (1/ 32) 106 i.e., 2  6105  16 1010  0 i.e.,   2   6105 10 105   5 rad i.e., 8 10 . 2 s Problem 10. A current of 4 A flows in a coil when connected to a 12 V dc source. If the same coil is connected to a 12 V, 50 rad/s ac source a current of 2.4 A flows in the circuit. Determine the inductance of the coil. Also find the power developed in the circuit if a 2500 F capacitor is connected in series with the coil. Solution: In case of a coil as Z  i.e., I  V  V Z so when dc is applied as   0 I  V i.e., R  12  3 … (i) R 4 and when ac is applied, I  V Z i.e., Z   V    12   5      or, R2  X2  52 (as Z  ) so, X2  52  R 2  52  32  42 i.e., X  4 L L but as X  L, L  XL  4 L  50  0.08  Now when the capacitor is connected to the above circuit in series, 1 1 103 As XC  C  50  2500 106  125  8  So, Z    5  and hence I  V  12  2.4 A Z 5 so, P  V I cos  (I  Z)  I  R  av rms rms rms rms     i.e., P  I2 R  (2.4)2  3  17.28 W . av rms Problem 11: A box contains L, C and R. When 250 V dc is applied to the terminals of the box, a current of 1.0 A flows in the circuit. When an ac source of 250 V rms at 2250 rad/s is connected, a current of 1.25 A rms flows. It is observed that the current rises with frequency and be- comes maximum at 4500 rad/s. Find the values of L, C and R. Draw the circuit diagram. Solution: As the circuit works on dc so all the element L, C and R cannot be in series. Further, as with change in frequency current increases, reaches a maximum and then decreases, which happens in series resonance, so L and C must be in series. So the box contains the circuit as shown in figure. Now when dc is applied, R V = IR V 250 i.e., R    I 1 so, R = 250  …(i) Now for ac as current is maximum at 4500 rad/s, 0  and as for ac, , i.e., LC  1 1 2 (4500)2 …(ii) I  V sin t  250 sin t  1sin t R R 250 I  V sin t X X so, I  I R  IX  1sin t  V sin t X or, with, I  I0 sin(t  ) I cos   1 and I sin   V 0 0  V 2 X  250 2 i.e., I2  1   or (1.25)2  1       i.e., X2  250  250 0.5625 or X  250 i.e., X  250  1000  0.75 3 i.e., L □ 1  1000 as X  L □ 1  C 3  C  or, 2250L □ 1    1000 …(iii) 2250C 3 as   2250 rad   s  Substituting the value of L in terms of C from equations (ii) and (iii), we have, 2250 1 C(4500)2 □ 1  1000 2250C 3 1  or, 2250 □ 1   1000 C  4500 4500 2250  3 1  1 or, □ 1   1000 C  9000 2250  3 or, 1  3  1000 i.e., C  106 F  1F C 9000 3 and so from Equation (ii), L  1  1  4  0.049 H . C2 106  (4500)2 81