Tiêu đề চল তড়িৎ -Written-Practice-Sheet_With-Solve.pdf ✅

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

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)