Tiêu đề 3. P2C3-চল তড়িৎ (Current Electricity) With Solve_Sha_5.12.23 PDF.pdf ✅

Pj Zwor  Varsity Practice Content 1 Z...Zxq Aa ̈vq Pj Zwor Current Electricity ACS Physics Department Gi g‡bvbxZ wjwLZ cÖkœmg~n 1| wb‡¤œi eZ©bxi g~j Zwor cÖevn wbY©q K‡iv| [Medium] 8 12 24V 4 6 mgvavb: 8 12 24V 4 6 A B B A 8 B 4 6 12 A 4 8 3  Rp1 = 4 × 8 4 + 8 = 2 3 × 4 = 8 3  Rp2 = 6 × 12 6 + 12 = 2 3 × 6 = 4   Req = 4 × 8 3 4 + 8 3 = 1.6   I = V R = 24 1.6 = 15 A (Ans.) 2| GKB AvK...wZwewkó `ywU Zv‡ii †ivav1⁄4 h_vμ‡g 1 I 2; Zv‡`i cÖ_‡g †kÖwY mgev‡q Ges c‡i mgvšÍivj mgev‡q hy3 K‡i Rs I Rp gv‡bi `ywU †iva MVb Kiv n‡jv| Rp : Rs Gi gvb wbY©q K‡iv| [Medium] mgvavb: R1 = k 1 , R2 = k 2  Rs = R1 + R2 = k(1 + 2)  Rp = R1R2 R1 + R2 = k 2 12 k(1 + 2) = k12 1 + 2  Rp : Rs = 12 1 + 2 × 1 1 + 2  Rp : Rs = 12 (1 + 2) 2 (Ans.) 3| wP‡Î I cÖevngvÎv wbY©q K‡ivÑ [Easy] 5 10 30 10 I 22.5V mgvavb: Req = 15 + 300 40 = 15 + 7.5 = 22.5 Ω  I = V Req = 1A  I30Ω = 10 10+30  1 = 1 4 = 0.25A (Ans.) 4| mylg cÖ ̄’‡”Q‡`i GKwU Zv‡ii †iva 9 Ω| ZviwU‡K mgvb 3 L‡Ð fvM Kiv n‡jv| LÐ wZbwUi •`N© ̈ 3 ̧Y Kiv n‡jv Ges mgvšÍivj mgev‡q hy3 Kiv n‡jv| mgevqwUi Zzj ̈ †iva? [Easy] mgvavb: 27 27 27 R1 = 9 3 = 3 Ω R2 = n2R1 = 32  3 = 27 Ω  R3 = 27 3 = 9 Ω (Ans.) 5| cÖ`Ë eZ©bx‡Z A I B we›`yi mv‡c‡ÿ Zzj ̈ Zwo”PvjK ej I Zzj ̈ Af ̈šÍixY ‡iva KZ? [Hard] 2 10V B 6V 4V A 1 2 mgvavb: Eeq1 = E1 r1 + E2 r2 1 r1 + 1 r2 = 10 2 + 4 2 1 2 + 1 2  Eeq1 = 5 + 2 1 = 7 V  Eeq2 = 7 – 6 = 1 V  Req = 2 2 + 1 = 2 Ω (Ans.)
2  Physics 2nd Paper Chapter-3 6| A I B we›`yi g‡a ̈ Zzj ̈‡iva wbY©q K‡iv| [Easy] A 5 5 5 5 5 5 B mgvavb: A 5 5 5 2  5 5 B A 5 3 5 B Rp = 5  15 2 5 + 15 2 = 3 Ω  Req = 13 Ω (Ans.) 7| A I B we›`yi g‡a ̈ Zzj ̈‡iva wbY©q K‡iv| [Medium] A B 8 4 4 4 4 8 mgvavb: 8 4 4 8 4 4  Req = 8 4 = 2 Ω (Ans.) 8| GKB AvKv‡ii `ywU avZe Zvi‡K †kÖwY mgev‡q hy3 Kiv n‡jv| hw` `ywU Zv‡ii Zwor cwievwnZv1⁄4 1 I 2 nq, Z‡e mgev‡qi Zzj ̈ cwievwnZv1⁄4? [Medium] mgvavb:  = 1   R = R1 + R2   2L A = 1 L A + 2 L A  2 = 1 + 2  2  = 1 1 + 1 2   = 212 1 + 2 (Ans.) 9| cÖwZwU †iv‡ai ga ̈ w`‡q Zwor cÖev‡ni gvbÑ [Medium] 2V 2V 2V 2V 2V 2V 1 1 1 mgvavb: 0V 0V A 2V 2V 2V 2V 2V 2V –2V –4V –6V –2V –4V –6V 1 1 1 awi, A we›`yi wefe 0 V,  †Kv‡bv †iv‡ai ga ̈ w`‡qB Zwor cÖevwnZ n‡e bv|  I = 0 A (Ans.) 10| 0.1 m •`N© ̈wewkó GKwU Zv‡i 5 V wefe cÖ‡f` cÖhy3 n‡jv| B‡j±a‡bi Zvob †eM 2.5  10–4 ms–1 | ZviwU‡Z B‡j±a‡bi NbZ¡ 8  1028 m –3 n‡j, ZviwUi Dcv`v‡bi Av‡cwÿK †iva? [Medium] mgvavb: I = V R = V ρ . L A  ρ = VA IL = VA Avne  L  ρ = V vneL = 5 2.5  10–4  8  1028  1.6  10–19  0.1  ρ = 50 20  1.6  105  ρ  5 2  3 2  10–5 = 5 3  10–5 = 5  0.3  10–5  ρ = 1.5  10–5 Ω m (Ans.) 11| A I B we›`yi ga ̈eZ©x wefe cv_©K ̈ wbY©q K‡iv| [Hard] A 3F 100V B A 1F 3F 1F 1F 20 10 mgvavb: aviK ̧‡jv PvwR©Z nIqvi ci eZ©bx‡Z †Kv‡bv Zwor cÖevwnZ n‡e bv| A 100V B A 3F 1F A C C C A C B B
Pj Zwor  Varsity Practice Content 3  I = 0  VAC = 100V aviK ̧‡jv‡K wfbœfv‡e mw3⁄4Z K‡i, A B B C A B 6F 2F 1F C  V6F = 2 2 + 6  100 [avi‡Ki Voltage Divider Law] = 2 8  100 = 25 V (Ans.) 12| GKB Dcv`v‡bi mgvb •`‡N© ̈i `ywU mylg Zv‡ii e ̈v‡mi AbycvZ 1 : 3| Zvi `ywU‡K †kÖwY mgev‡q hy3 K‡i G‡`i ga ̈ w`‡q GKwU wbw`©ó mg‡qi Rb ̈ Zwor cÖevn Pvjbv Kiv n‡j, Zvi `ywU‡Z Drcbœ Zv‡ci AbycvZ? [Easy] mgvavb: R =  L A =  L   d 2 4     r = d 2  R  1 d 2  H = I2Rt [I, t const]  H  R  1 d 2  H  1 d 2  H1 H2 =     d2 d1 2 =     3 1 2 = 9 : 1 (Ans.) 13| 100  †iv‡ai GKwU KzÐjxi ga ̈ w`‡q 3 A cÖevn 1 min a‡i Pvjbv Ki‡j 1 kg Zi‡ji ZvcgvÎv 30C e„w× cvq| Zi‡ji Av‡cwÿK Zvc wbY©q K‡iv| [Easy] mgvavb: H = ms  I 2Rt = ms  S = 3 2  100  60 1  30 = 9  100  2  S = 1800 Jkg–1K –1 (Ans.) 14| P I Q Zv‡ii •`‡N© ̈i AbycvZ 3 : 1, e ̈v‡mi AbycvZ 1 : 2 Ges †ivav‡1⁄4i AbycvZ 1:20| G‡`i‡K mgvšÍivj mgev‡q 2 V ZworPvjK ej Ges D‡cÿYxq Af ̈šÍixY †iva wewkó GKwU †Kv‡li mv‡_ hy3 Kiv n‡jv| GB `yB Zv‡i Drcbœ Zv‡ci AbycvZÑ [Medium] mgvavb: R =  L A =  L   d 2 4  R  L d 2  H = V 2 R t  H  1 R [V, t const]  H1 H2 = R2 R1 = 2 1 L2 L1      d1 d2 2  H1 H2 = 20  1 3      1 2 2  H1 H2 = 5 : 3 (Ans.) 15| wP‡Î cÖ`wk©Z cÖwZwU †iv‡ai m‡e©v”P ÿgZv 20 W| mgMÖ eZ©bxwU m‡e©v”P wK ÿgZvq kw3 e ̈q Ki‡e? [Medium] I R3 = 4 R1 = 4 R2 = 4 mgvavb: R3 Gi ga ̈ w`‡q m‡e©v”P Zwor cÖevn, P = I2R  I = P R = 20 4 = 5 A  eZ©bxi Zzj ̈ †iva, Req = 4 2 + 4 = 6  myZivs, mgMÖ eZ©bxi kw3 e ̈‡qi nvi, P = I 2 Req  P = ( 5) 2  6 = 30 W (Ans.) 16| wP‡Î cÖ`wk©Z wZbwU †ivaK R1, R2, R3 †Z cÖwZ wgwb‡U Drcbœ Zv‡ci AbycvZ? [Medium] R1 = 6 R2 = 3 12V R3 = 1 mgvavb: Rp = 2 3  3 = 2  V1 = V2 = 2 2 + 1  12 = 8 V  V3 = 12 – 8 = 4 V GLb, H = V 2 R t  H  V 2 R [t const]  H1 : H2 : H3 = V1 2 R1 : V2 2 R2 : V3 2 R3 = 8 2 6 : 8 2 3 : 4 2 1 = 32 3 : 64 3 : 16 = 32 : 64 : 48 H1 : H2 : H3 = 2 : 4 : 3 (Ans.)
4  Physics 2nd Paper Chapter-3 17| 20  Af ̈šÍixb †iv‡ai M ̈vjfv‡bvwgUv‡ii ga ̈ w`‡q 1 A ZworcÖevn Pj‡Q| GKwU kv›U e ̈env‡ii d‡j GB cÖevn K‡g 0.01 A nq| kv‡›Ui †iva KZ? [Easy] mgvavb: R = G n – 1 = 20 1 0.01 – 1  R = 20 99  (Ans.) 18| 100  †iv‡ai GKwU M ̈vjfv‡bvwgUvi 0.1 A Zwor cÖevn gvÎvq c~Y© † ̄‹j we‡ÿc †`q| 100 V G c~Y© † ̄‹j we‡ÿc †`q Giƒc GKwU †fvëwgUv‡i iƒcvšÍwiZ Ki‡Z KZ †iva †kÖwY mgev‡q hy3 Ki‡Z n‡e? [Easy] mgvavb: V = IR = 100  0.1 = 10 V †fvëwgUv‡ii cvjøv e„w×i Rb ̈, RS = (n – 1) RV  RS =     100 10 – 1  100  RS = 900  (Ans.) 19| 90  †iv‡ai GKwU M ̈vjfv‡bvwgUv‡ii mv‡_ 10  Gi kv›U e ̈envi Ki‡j g~j cÖev‡ni kZKiv KZ Ask M ̈vjfv‡bvwgUv‡ii ga ̈ w`‡q cÖevwnZ n‡e? [Easy] mgvavb: I G = 90 R = 10 IG = R R + G  I  IG I = 10 10 + 90  IG I  100% = 10 100  100%  IG I  100% = 10% (Ans.) 20| G  †iv‡ai †Kv‡bv M ̈vjfv‡bvwgUv‡ii mv‡_ GKwU mv›U hy3 Kivq M ̈vjfv‡bvwgUv‡ii wfZi w`‡q g~j cÖev‡ni 1 n Ask cÖevwnZ nq| cÖgvY K‡iv †h, mv‡›Ui †iva G n – 1 | [Easy] mgvavb: I I n G R eZ©bx n‡Z, I n = R G + R  I  n = G + R R  G R + 1 = n  R = G n – 1 (Ans.) 21| 20  †iv‡ai GKwU M ̈vjfv‡bvwgUv‡ii mv‡_ KZ †iv‡ai GKwU mv›U hy3 Ki‡j †gvU Zwor cÖevngvÎvi 1% M ̈vjfv‡bvwgUv‡ii ga ̈ w`‡q hv‡e? [Easy] mgvavb: 1%  I = R G + R  I  1 100 = R G + R  G R = 100 – 1 = 99  R = G 99 = 20 99  (Ans.) 22| wP‡Îi eZ©bx‡Z a, b I c we›`yi wefe h_vμ‡g 30 V, 12 V I 2 V| wZbwU †iva‡Ki Zwor cÖevn wbY©q K‡iv| [Medium] 10 30 20 a c b mgvavb: 10 30 20 a c I1 b I2 O V I awi, O we›`yi wefe V0  I = I1 + I2  Va – V0 10 = V0 – Vb 20 + V0 – Vc 30  30 – V0 10 = V0 – 12 20 + V0 – 2 30  180 – 6V0 = 3V0 – 36 + 2V0 – 4  11V0 = 220  V0 = 20 V  I = 30 – 20 10 = 1 A I1 = 30 30 + 20  1 = 3 5 A I2 = 20 20 + 30  1 = 2 5 A (Ans.) 23| wb‡¤œi eZ©bxi cÖwZwU kvLvq Zwor cÖevn wbY©q K‡iv| [Medium] 5V 20 20V 5 10 mgvavb: I1 20 20V 5 10 5V I2 cÖ_g jy‡c, 5I1 + 20(I1 – I2) – 5 = 0  5I1 – 4I2 – 1 = 0 ........... (i) wØZxq jy‡c, 10I2 + 20 + 20(I2 – I1) = 0  – 2I1 + 3I2 + 2 = 0 ........... (ii)