Page 1 Introductory Exercise 25.1 1. R V I = = = DC 100 10 10W Z V I = = = AC 150 10 15W X Z R L = - 2 2 = - ( ) ( ) 15 10 2 2 =5 5W L X X f L L = = w p 2 = ´ ´ 5 5 2 50 3.14 »0.036 H \ V IX L L = = 50 5 V =111.8 V 2. For phase an gle to be zero, X X L C = Þ w w L C = 1 Þ L C f C = = 1 1 2 2 2 w p ( ) = ´ - 1 360 10 2 6 ( ) »7.7 H As X X L C = \ Z R = I V Z = = = 120 20 6 A In tro duc tory Ex er cise 25.2 1. Res o nat ing fre quency, w r L C = = ´ ´ - 1 1 2 1 0 6 0 . 0 3 = 1 0 6 4 f r r = = ´ ´ w p 2 10 2 6 4 3.14 » 1105 Hz Phase angle at resonance is always 0°. 2. Re sis tance of arc lamp, R V I = DC = = 40 10 4W Impedance of series combination, Z V I = = = AC 200 10 20W Power factor = f = = cos R Z 4 20 = 1 5 Alternating Current 25 Page 2 Introductory Exercise 25.1 1. R V I = = = DC 100 10 10W Z V I = = = AC 150 10 15W X Z R L = - 2 2 = - ( ) ( ) 15 10 2 2 =5 5W L X X f L L = = w p 2 = ´ ´ 5 5 2 50 3.14 »0.036 H \ V IX L L = = 50 5 V =111.8 V 2. For phase an gle to be zero, X X L C = Þ w w L C = 1 Þ L C f C = = 1 1 2 2 2 w p ( ) = ´ - 1 360 10 2 6 ( ) »7.7 H As X X L C = \ Z R = I V Z = = = 120 20 6 A In tro duc tory Ex er cise 25.2 1. Res o nat ing fre quency, w r L C = = ´ ´ - 1 1 2 1 0 6 0 . 0 3 = 1 0 6 4 f r r = = ´ ´ w p 2 10 2 6 4 3.14 » 1105 Hz Phase angle at resonance is always 0°. 2. Re sis tance of arc lamp, R V I = DC = = 40 10 4W Impedance of series combination, Z V I = = = AC 200 10 20W Power factor = f = = cos R Z 4 20 = 1 5 Alternating Current 25 AIEEE Corner Sub jec ti ve Ques tions (Level-1) 1. (a) X L fL L = = w p 2 = ´ ´ ´ 2 50 2 3.14 =628W (b) X L L =w Þ L X L = w = = ´ ´ X f L 2 2 2 50 p 3.14 = 6.37 mH (c) X C fC C = = 1 1 2 w p = ´ ´ ´ ´ - 1 2 50 2 10 6 3.14 = = 1592 1.59 k W W (d) X C C = 1 w Þ C X C = 1 w = ´ ´ ´ = 1 2 50 2 3.14 1.59mF 2. (a) Z R X X L C = + - 2 2 ( ) = + - æ è ç ç ö ø ÷ ÷ R L C 2 2 1 w w = + ´ - ´ ´ æ è ç ç ö ø ÷ ÷ - ( ) 300 400 1 400 8 10 2 6 2 0.25 =367.6W I V Z 0 0 120 = = = 367.6 0.326 A (b) f = - - tan 1 X X R L C = » - ° - tan 1 300 212.5 35.3 As X X C L > voltage will lag behind current by 35.3°. (c) V I R R = = ´ 0 300 0.326 = 97.8 V, V I X L L = = 0 32.6V V I X C C = = ´ 0 0.326 312.5 = » 101.875 V 120 V 3. (a) Power fac tor at res o nance is al ways 1, as Z R = , Power factor = f = = cos R Z 1. (b) P I E E R = f = 0 0 0 2 2 2 cos = ´ = ( ) 150 3 150 75 2 W (c) Because resonance is still maintained, average power consumed will remain same, i.e., 75 W. 4. (a) As voltage is lag behind current, inductor should be added to the circuit to raise the power factor. (b) Power factor = f = cos R Z Þ Z R = f = cos 60 0.720 = 250 3 W X Z R C = - 2 2 = æ è ç ö ø ÷ - 250 3 60 2 2 ( ) =58W C X C = 1 w = 1 2pf X C = ´ ´ ´ 1 2 50 58 3.14 =54mF For resonance, w r LC = 1 Þ L C r = 1 2 w = 1 2 2 ( ) pf C Þ L = ´ ´ ´ ´ - 1 2 50 54 10 2 6 ( ) 3.14 =0.185 H 131 Page 3 Introductory Exercise 25.1 1. R V I = = = DC 100 10 10W Z V I = = = AC 150 10 15W X Z R L = - 2 2 = - ( ) ( ) 15 10 2 2 =5 5W L X X f L L = = w p 2 = ´ ´ 5 5 2 50 3.14 »0.036 H \ V IX L L = = 50 5 V =111.8 V 2. For phase an gle to be zero, X X L C = Þ w w L C = 1 Þ L C f C = = 1 1 2 2 2 w p ( ) = ´ - 1 360 10 2 6 ( ) »7.7 H As X X L C = \ Z R = I V Z = = = 120 20 6 A In tro duc tory Ex er cise 25.2 1. Res o nat ing fre quency, w r L C = = ´ ´ - 1 1 2 1 0 6 0 . 0 3 = 1 0 6 4 f r r = = ´ ´ w p 2 10 2 6 4 3.14 » 1105 Hz Phase angle at resonance is always 0°. 2. Re sis tance of arc lamp, R V I = DC = = 40 10 4W Impedance of series combination, Z V I = = = AC 200 10 20W Power factor = f = = cos R Z 4 20 = 1 5 Alternating Current 25 AIEEE Corner Sub jec ti ve Ques tions (Level-1) 1. (a) X L fL L = = w p 2 = ´ ´ ´ 2 50 2 3.14 =628W (b) X L L =w Þ L X L = w = = ´ ´ X f L 2 2 2 50 p 3.14 = 6.37 mH (c) X C fC C = = 1 1 2 w p = ´ ´ ´ ´ - 1 2 50 2 10 6 3.14 = = 1592 1.59 k W W (d) X C C = 1 w Þ C X C = 1 w = ´ ´ ´ = 1 2 50 2 3.14 1.59mF 2. (a) Z R X X L C = + - 2 2 ( ) = + - æ è ç ç ö ø ÷ ÷ R L C 2 2 1 w w = + ´ - ´ ´ æ è ç ç ö ø ÷ ÷ - ( ) 300 400 1 400 8 10 2 6 2 0.25 =367.6W I V Z 0 0 120 = = = 367.6 0.326 A (b) f = - - tan 1 X X R L C = » - ° - tan 1 300 212.5 35.3 As X X C L > voltage will lag behind current by 35.3°. (c) V I R R = = ´ 0 300 0.326 = 97.8 V, V I X L L = = 0 32.6V V I X C C = = ´ 0 0.326 312.5 = » 101.875 V 120 V 3. (a) Power fac tor at res o nance is al ways 1, as Z R = , Power factor = f = = cos R Z 1. (b) P I E E R = f = 0 0 0 2 2 2 cos = ´ = ( ) 150 3 150 75 2 W (c) Because resonance is still maintained, average power consumed will remain same, i.e., 75 W. 4. (a) As voltage is lag behind current, inductor should be added to the circuit to raise the power factor. (b) Power factor = f = cos R Z Þ Z R = f = cos 60 0.720 = 250 3 W X Z R C = - 2 2 = æ è ç ö ø ÷ - 250 3 60 2 2 ( ) =58W C X C = 1 w = 1 2pf X C = ´ ´ ´ 1 2 50 58 3.14 =54mF For resonance, w r LC = 1 Þ L C r = 1 2 w = 1 2 2 ( ) pf C Þ L = ´ ´ ´ ´ - 1 2 50 54 10 2 6 ( ) 3.14 =0.185 H 131 5. V t t ( ) sin ( / ) = + 170 6280 3 p volt i t t ( ) sin ( / ) = + 8.5 6280 2 p amp. (b) f = = ´ = w p 2 6280 2 1000 3.14 Hz =1 kHz (c) f = - = p p p 2 3 6 Þ cos cos f = = p 6 3 2 As phase of i is greater than V, current is leading voltage. (d)Clearly the circuit is capacitive in nature, we have cos f = R Z Þ 3 2 = R Z Þ Z R = 2 3 Also, Z V i = = = 0 0 170 20 8.5 W R Z = = 3 2 10 3 W Again, Z R X C = + 2 2 Þ X Z R C = - 2 2 = - = 400 300 10W X C X C C = Þ = ´ 1 1 1 6280 10 w w =1592 . mF 6. I V X V L L = = w (a) w=100 rad/s \ I = ´ = 60 100 5 0.12 A (b) w = 1000 rad/s \ I = ´ = ´ - 60 1000 5 10 2 1.2 A (c) w = 10000 rad/s \ I = ´ = ´ - 60 10000 5 10 3 1.2 A 7. V t R = ( )cos [( ) ] 2.5 V rad/s 950 (a) I V R R = = ( ) cos [( 2.5 V rad/s) ] 950 300 t =( cos[( ) ] 8. mA) rad/s 33 950 t (b) X L L = = ´ w 950 0.800 =760W (c) V I X t L L = + 0 2 cos( / ) w p Þ V I X t L L = - 0 sin w = - 6.33 rad / s sin [( ) ] 950 t V 8. Given, L = 0.120 H, R =240W, C =7.30 F m , I rms .450 A =0 , f = 400 Hz X L fL L = = w p 2 = ´ ´ ´ 2 400 3.14 0.120 =301.44W X C f C C = = 1 1 2 w p = ´ ´ ´ ´ - 1 2 400 10 6 3.14 7.3 =54.43W (a) cos ( ) f= = + - R Z R R X X L C 2 2 = + - 240 240 2 2 ( ) ( ) 301.44 54.43 =0.697 f= » ° - cos ( . ) . 1 0697 458 (b) Z R X X L C = + - 2 2 ( ) = + - ( ) ( ) 240 2 2 301.44 54.43 =344 W (c) V I Z rms rms 0.450 = = ´ 344 =154.8 V »155 V (d) P V I av = f rms rms cos = ´ ´ = 155 0.450 0.697 48.6W (e) P I R R = = ´ rms 0.450) 2 2 240 ( = 48.6 W (f) and (g) Average power associated with inductor and capacitor is always zero. 132 170V V â€“170V O 0.25 0.50 0.75 1.00 1.25 t (ms) i O 1 â€” 12 1 â€” 3 7 â€” 12 5 â€” 6 13 â€” 12 4 â€” 3 t (ms) Page 4 Introductory Exercise 25.1 1. R V I = = = DC 100 10 10W Z V I = = = AC 150 10 15W X Z R L = - 2 2 = - ( ) ( ) 15 10 2 2 =5 5W L X X f L L = = w p 2 = ´ ´ 5 5 2 50 3.14 »0.036 H \ V IX L L = = 50 5 V =111.8 V 2. For phase an gle to be zero, X X L C = Þ w w L C = 1 Þ L C f C = = 1 1 2 2 2 w p ( ) = ´ - 1 360 10 2 6 ( ) »7.7 H As X X L C = \ Z R = I V Z = = = 120 20 6 A In tro duc tory Ex er cise 25.2 1. Res o nat ing fre quency, w r L C = = ´ ´ - 1 1 2 1 0 6 0 . 0 3 = 1 0 6 4 f r r = = ´ ´ w p 2 10 2 6 4 3.14 » 1105 Hz Phase angle at resonance is always 0°. 2. Re sis tance of arc lamp, R V I = DC = = 40 10 4W Impedance of series combination, Z V I = = = AC 200 10 20W Power factor = f = = cos R Z 4 20 = 1 5 Alternating Current 25 AIEEE Corner Sub jec ti ve Ques tions (Level-1) 1. (a) X L fL L = = w p 2 = ´ ´ ´ 2 50 2 3.14 =628W (b) X L L =w Þ L X L = w = = ´ ´ X f L 2 2 2 50 p 3.14 = 6.37 mH (c) X C fC C = = 1 1 2 w p = ´ ´ ´ ´ - 1 2 50 2 10 6 3.14 = = 1592 1.59 k W W (d) X C C = 1 w Þ C X C = 1 w = ´ ´ ´ = 1 2 50 2 3.14 1.59mF 2. (a) Z R X X L C = + - 2 2 ( ) = + - æ è ç ç ö ø ÷ ÷ R L C 2 2 1 w w = + ´ - ´ ´ æ è ç ç ö ø ÷ ÷ - ( ) 300 400 1 400 8 10 2 6 2 0.25 =367.6W I V Z 0 0 120 = = = 367.6 0.326 A (b) f = - - tan 1 X X R L C = » - ° - tan 1 300 212.5 35.3 As X X C L > voltage will lag behind current by 35.3°. (c) V I R R = = ´ 0 300 0.326 = 97.8 V, V I X L L = = 0 32.6V V I X C C = = ´ 0 0.326 312.5 = » 101.875 V 120 V 3. (a) Power fac tor at res o nance is al ways 1, as Z R = , Power factor = f = = cos R Z 1. (b) P I E E R = f = 0 0 0 2 2 2 cos = ´ = ( ) 150 3 150 75 2 W (c) Because resonance is still maintained, average power consumed will remain same, i.e., 75 W. 4. (a) As voltage is lag behind current, inductor should be added to the circuit to raise the power factor. (b) Power factor = f = cos R Z Þ Z R = f = cos 60 0.720 = 250 3 W X Z R C = - 2 2 = æ è ç ö ø ÷ - 250 3 60 2 2 ( ) =58W C X C = 1 w = 1 2pf X C = ´ ´ ´ 1 2 50 58 3.14 =54mF For resonance, w r LC = 1 Þ L C r = 1 2 w = 1 2 2 ( ) pf C Þ L = ´ ´ ´ ´ - 1 2 50 54 10 2 6 ( ) 3.14 =0.185 H 131 5. V t t ( ) sin ( / ) = + 170 6280 3 p volt i t t ( ) sin ( / ) = + 8.5 6280 2 p amp. (b) f = = ´ = w p 2 6280 2 1000 3.14 Hz =1 kHz (c) f = - = p p p 2 3 6 Þ cos cos f = = p 6 3 2 As phase of i is greater than V, current is leading voltage. (d)Clearly the circuit is capacitive in nature, we have cos f = R Z Þ 3 2 = R Z Þ Z R = 2 3 Also, Z V i = = = 0 0 170 20 8.5 W R Z = = 3 2 10 3 W Again, Z R X C = + 2 2 Þ X Z R C = - 2 2 = - = 400 300 10W X C X C C = Þ = ´ 1 1 1 6280 10 w w =1592 . mF 6. I V X V L L = = w (a) w=100 rad/s \ I = ´ = 60 100 5 0.12 A (b) w = 1000 rad/s \ I = ´ = ´ - 60 1000 5 10 2 1.2 A (c) w = 10000 rad/s \ I = ´ = ´ - 60 10000 5 10 3 1.2 A 7. V t R = ( )cos [( ) ] 2.5 V rad/s 950 (a) I V R R = = ( ) cos [( 2.5 V rad/s) ] 950 300 t =( cos[( ) ] 8. mA) rad/s 33 950 t (b) X L L = = ´ w 950 0.800 =760W (c) V I X t L L = + 0 2 cos( / ) w p Þ V I X t L L = - 0 sin w = - 6.33 rad / s sin [( ) ] 950 t V 8. Given, L = 0.120 H, R =240W, C =7.30 F m , I rms .450 A =0 , f = 400 Hz X L fL L = = w p 2 = ´ ´ ´ 2 400 3.14 0.120 =301.44W X C f C C = = 1 1 2 w p = ´ ´ ´ ´ - 1 2 400 10 6 3.14 7.3 =54.43W (a) cos ( ) f= = + - R Z R R X X L C 2 2 = + - 240 240 2 2 ( ) ( ) 301.44 54.43 =0.697 f= » ° - cos ( . ) . 1 0697 458 (b) Z R X X L C = + - 2 2 ( ) = + - ( ) ( ) 240 2 2 301.44 54.43 =344 W (c) V I Z rms rms 0.450 = = ´ 344 =154.8 V »155 V (d) P V I av = f rms rms cos = ´ ´ = 155 0.450 0.697 48.6W (e) P I R R = = ´ rms 0.450) 2 2 240 ( = 48.6 W (f) and (g) Average power associated with inductor and capacitor is always zero. 132 170V V â€“170V O 0.25 0.50 0.75 1.00 1.25 t (ms) i O 1 â€” 12 1 â€” 3 7 â€” 12 5 â€” 6 13 â€” 12 4 â€” 3 t (ms) Objective Questions (Level-1) 1. In an AC cir cuit, cosf is called power fac tor. 2. DC am me ter mea sures charge flow ing in the cir cuit per unit time, hence it mea sures av er age value of cur rent, but av er age value of AC over a long time is zero. 3. Z R X X L C = + - 2 2 ( ) = + - æ è ç ç ö ø ÷ ÷ R L C 2 2 1 w w Hence, for X X L C < , Z decreases with increase in frequency and for X X L C > , Z increases with increase in frequency. 4. As volt age leads cur rent and f < p 2 , hence ei ther cir cuit con tain s in du c tance and re sis tance or con tain s in du c tance, ca pac i tan ce and re sis tance with X X L C > . 5. RMS value of sine wave AC is 0.707I 0 , but can be dif fer ent for dif fer ent types of ACâ€™s. 6. P I E v v = f= cos 0 7. Z R X X L C = + - 2 2 ( ) 8. P V I = 0 0 2 [V 0 and I 0 are peak volt age and cur rent through re sis tor only] 9. V V rms = = 0 2 170 V f = = ´ w p 2 120 2 3.14 »19 Hz 10. Cur r ent is max i mum at w w = = r L C 1 = ´ ´ - 1 8 10 6 0.5 =500 rad/s. 11. P I E = f 0 0 2 cos = ´ ´ = - 100 100 2 3 10 3 cos p 2.5W 12. X C C = =¥ 1 w if w = 0, i.e., for DC 13. V t = 10 100 cos p at t = 1 600 s, V =10 100 1 600 cos p = = ´ = 10 6 10 3 2 5 3 cos p V 14. For purely re sis tive cir cuit f = 0. 15. X C C = 1 w Þ X C µ 1 w or X f C µ 1 16. sinf= = X Z 1 3 Þ f= é ë ê ù û ú - sin 1 1 3 17. f= = f= 3 2 2 0 0 0 p , cos P I E 18. R V I = = DC DC 100W Z V I = = = AC AC 0.5 100 200W X Z R L = - = 2 2 100 3 W L X X f L L = = = ´ w p p 2 100 3 2 50 = æ è ç ç ö ø ÷ ÷ 3 p H 19. I V X CV C rms rms rms = = w = ´ ´ ´ - 100 1 10 200 2 2 6 I rms mA =20 20. V V V R L = + = + 2 2 2 2 20 15 ( ) ( ) =25 V, V 0 25 2 = V 21. P I V = f = 0 0 2 0 cos Þ cos f =0 Þ f= ° 90 22. R is in de pend ent of fre quency. 23. L is very high so that cir cuit con sumes less power. 133 Page 5 Introductory Exercise 25.1 1. R V I = = = DC 100 10 10W Z V I = = = AC 150 10 15W X Z R L = - 2 2 = - ( ) ( ) 15 10 2 2 =5 5W L X X f L L = = w p 2 = ´ ´ 5 5 2 50 3.14 »0.036 H \ V IX L L = = 50 5 V =111.8 V 2. For phase an gle to be zero, X X L C = Þ w w L C = 1 Þ L C f C = = 1 1 2 2 2 w p ( ) = ´ - 1 360 10 2 6 ( ) »7.7 H As X X L C = \ Z R = I V Z = = = 120 20 6 A In tro duc tory Ex er cise 25.2 1. Res o nat ing fre quency, w r L C = = ´ ´ - 1 1 2 1 0 6 0 . 0 3 = 1 0 6 4 f r r = = ´ ´ w p 2 10 2 6 4 3.14 » 1105 Hz Phase angle at resonance is always 0°. 2. Re sis tance of arc lamp, R V I = DC = = 40 10 4W Impedance of series combination, Z V I = = = AC 200 10 20W Power factor = f = = cos R Z 4 20 = 1 5 Alternating Current 25 AIEEE Corner Sub jec ti ve Ques tions (Level-1) 1. (a) X L fL L = = w p 2 = ´ ´ ´ 2 50 2 3.14 =628W (b) X L L =w Þ L X L = w = = ´ ´ X f L 2 2 2 50 p 3.14 = 6.37 mH (c) X C fC C = = 1 1 2 w p = ´ ´ ´ ´ - 1 2 50 2 10 6 3.14 = = 1592 1.59 k W W (d) X C C = 1 w Þ C X C = 1 w = ´ ´ ´ = 1 2 50 2 3.14 1.59mF 2. (a) Z R X X L C = + - 2 2 ( ) = + - æ è ç ç ö ø ÷ ÷ R L C 2 2 1 w w = + ´ - ´ ´ æ è ç ç ö ø ÷ ÷ - ( ) 300 400 1 400 8 10 2 6 2 0.25 =367.6W I V Z 0 0 120 = = = 367.6 0.326 A (b) f = - - tan 1 X X R L C = » - ° - tan 1 300 212.5 35.3 As X X C L > voltage will lag behind current by 35.3°. (c) V I R R = = ´ 0 300 0.326 = 97.8 V, V I X L L = = 0 32.6V V I X C C = = ´ 0 0.326 312.5 = » 101.875 V 120 V 3. (a) Power fac tor at res o nance is al ways 1, as Z R = , Power factor = f = = cos R Z 1. (b) P I E E R = f = 0 0 0 2 2 2 cos = ´ = ( ) 150 3 150 75 2 W (c) Because resonance is still maintained, average power consumed will remain same, i.e., 75 W. 4. (a) As voltage is lag behind current, inductor should be added to the circuit to raise the power factor. (b) Power factor = f = cos R Z Þ Z R = f = cos 60 0.720 = 250 3 W X Z R C = - 2 2 = æ è ç ö ø ÷ - 250 3 60 2 2 ( ) =58W C X C = 1 w = 1 2pf X C = ´ ´ ´ 1 2 50 58 3.14 =54mF For resonance, w r LC = 1 Þ L C r = 1 2 w = 1 2 2 ( ) pf C Þ L = ´ ´ ´ ´ - 1 2 50 54 10 2 6 ( ) 3.14 =0.185 H 131 5. V t t ( ) sin ( / ) = + 170 6280 3 p volt i t t ( ) sin ( / ) = + 8.5 6280 2 p amp. (b) f = = ´ = w p 2 6280 2 1000 3.14 Hz =1 kHz (c) f = - = p p p 2 3 6 Þ cos cos f = = p 6 3 2 As phase of i is greater than V, current is leading voltage. (d)Clearly the circuit is capacitive in nature, we have cos f = R Z Þ 3 2 = R Z Þ Z R = 2 3 Also, Z V i = = = 0 0 170 20 8.5 W R Z = = 3 2 10 3 W Again, Z R X C = + 2 2 Þ X Z R C = - 2 2 = - = 400 300 10W X C X C C = Þ = ´ 1 1 1 6280 10 w w =1592 . mF 6. I V X V L L = = w (a) w=100 rad/s \ I = ´ = 60 100 5 0.12 A (b) w = 1000 rad/s \ I = ´ = ´ - 60 1000 5 10 2 1.2 A (c) w = 10000 rad/s \ I = ´ = ´ - 60 10000 5 10 3 1.2 A 7. V t R = ( )cos [( ) ] 2.5 V rad/s 950 (a) I V R R = = ( ) cos [( 2.5 V rad/s) ] 950 300 t =( cos[( ) ] 8. mA) rad/s 33 950 t (b) X L L = = ´ w 950 0.800 =760W (c) V I X t L L = + 0 2 cos( / ) w p Þ V I X t L L = - 0 sin w = - 6.33 rad / s sin [( ) ] 950 t V 8. Given, L = 0.120 H, R =240W, C =7.30 F m , I rms .450 A =0 , f = 400 Hz X L fL L = = w p 2 = ´ ´ ´ 2 400 3.14 0.120 =301.44W X C f C C = = 1 1 2 w p = ´ ´ ´ ´ - 1 2 400 10 6 3.14 7.3 =54.43W (a) cos ( ) f= = + - R Z R R X X L C 2 2 = + - 240 240 2 2 ( ) ( ) 301.44 54.43 =0.697 f= » ° - cos ( . ) . 1 0697 458 (b) Z R X X L C = + - 2 2 ( ) = + - ( ) ( ) 240 2 2 301.44 54.43 =344 W (c) V I Z rms rms 0.450 = = ´ 344 =154.8 V »155 V (d) P V I av = f rms rms cos = ´ ´ = 155 0.450 0.697 48.6W (e) P I R R = = ´ rms 0.450) 2 2 240 ( = 48.6 W (f) and (g) Average power associated with inductor and capacitor is always zero. 132 170V V â€“170V O 0.25 0.50 0.75 1.00 1.25 t (ms) i O 1 â€” 12 1 â€” 3 7 â€” 12 5 â€” 6 13 â€” 12 4 â€” 3 t (ms) Objective Questions (Level-1) 1. In an AC cir cuit, cosf is called power fac tor. 2. DC am me ter mea sures charge flow ing in the cir cuit per unit time, hence it mea sures av er age value of cur rent, but av er age value of AC over a long time is zero. 3. Z R X X L C = + - 2 2 ( ) = + - æ è ç ç ö ø ÷ ÷ R L C 2 2 1 w w Hence, for X X L C < , Z decreases with increase in frequency and for X X L C > , Z increases with increase in frequency. 4. As volt age leads cur rent and f < p 2 , hence ei ther cir cuit con tain s in du c tance and re sis tance or con tain s in du c tance, ca pac i tan ce and re sis tance with X X L C > . 5. RMS value of sine wave AC is 0.707I 0 , but can be dif fer ent for dif fer ent types of ACâ€™s. 6. P I E v v = f= cos 0 7. Z R X X L C = + - 2 2 ( ) 8. P V I = 0 0 2 [V 0 and I 0 are peak volt age and cur rent through re sis tor only] 9. V V rms = = 0 2 170 V f = = ´ w p 2 120 2 3.14 »19 Hz 10. Cur r ent is max i mum at w w = = r L C 1 = ´ ´ - 1 8 10 6 0.5 =500 rad/s. 11. P I E = f 0 0 2 cos = ´ ´ = - 100 100 2 3 10 3 cos p 2.5W 12. X C C = =¥ 1 w if w = 0, i.e., for DC 13. V t = 10 100 cos p at t = 1 600 s, V =10 100 1 600 cos p = = ´ = 10 6 10 3 2 5 3 cos p V 14. For purely re sis tive cir cuit f = 0. 15. X C C = 1 w Þ X C µ 1 w or X f C µ 1 16. sinf= = X Z 1 3 Þ f= é ë ê ù û ú - sin 1 1 3 17. f= = f= 3 2 2 0 0 0 p , cos P I E 18. R V I = = DC DC 100W Z V I = = = AC AC 0.5 100 200W X Z R L = - = 2 2 100 3 W L X X f L L = = = ´ w p p 2 100 3 2 50 = æ è ç ç ö ø ÷ ÷ 3 p H 19. I V X CV C rms rms rms = = w = ´ ´ ´ - 100 1 10 200 2 2 6 I rms mA =20 20. V V V R L = + = + 2 2 2 2 20 15 ( ) ( ) =25 V, V 0 25 2 = V 21. P I V = f = 0 0 2 0 cos Þ cos f =0 Þ f= ° 90 22. R is in de pend ent of fre quency. 23. L is very high so that cir cuit con sumes less power. 133 24. tan f = X R L Þ tan 45 100 ° = X L Þ X L =100W wL =100W L = ´ ´ » 100 2 10 16 3 3.14 mH 25. The min i mum time taken by it in reach ing from zero to peak value = T 4 = = ´ = = 1 4 1 4 50 1 200 5 f ms 26. f = ° 60 P I V = f = ´ ´ 0 0 2 4 220 1 2 2 cos =220 W JEE Corner Assertion and Reasons 1. X C and X L can be greater than Z be cause Z R X X L C = + - 2 2 ( ) Hence, V IX C C = and V IX L L = can be greater than V IZ = . 2. At res o nan ce X X L C = , with fur ther in crease in fre quency, X L in creases but X C de creases hence volt age will lead cur rent. 3. f L C r = 1 2p , if di elec tric slab is in serted be tween the plates of the ca pac i tor, its ca pac i tan ce will in crease, hence, f r will decrease. 4. q= Area under graph = ´ ´ + + ´ ´ + 1 2 4 2 3 1 2 4 2 4 ( ) ( ) =22C Average current = = = q t 22 6 3.6 A 5. On in sert in g fer ro mag n etic rod in sid e the in duc t or, X L and hence V L in crease. Due to this cur rent will in crease if it is lag ging and vice-versa. 6. V V V R L C = = Þ R X X L C = = Hence, f =0 and I is maximum. as Z R X X L C = + - 2 2 ( ) is minimum. 7. I I I L C = - = 0 8. P I R = = ´ = rms W 2 2 2 10 20 ( ) 9. In duc tor coil re sists vary in g cur rent. 10. I E R L 0 0 2 2 2 = + w , f = - tan 1 w L R 11. At res o nance, cur rent and volt age are in same phase and I V R 0 0 = . Hence, I 0 de pends on R . Objective Questions (Level-2) Sin gle Cor rect Op tions 1. For par al l el cir cuit f = é ë ê ù û ú - t a n / / 1 1 1 X R L = - t a n 1 4 3 = ° 53 2. Cur rent will re main same in se ries cir cuit given by I I t = -f 0 sin( ) w = - æ è ç ö ø ÷ - I t X R L 0 1 sin tan w 3. R R R L = + = 1 10 W X L L = = w 10 W, X C C = = 1 10 w W 134Read More

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