Tiêu đề 3. P2C3. HSC PREP Papers 26_With Solve.pdf ✅

Pj Zwor  HSC Prep Papers 1 Z...Zxq Aa ̈vq Pj Zwor Current Electricity Topicwise CQ Trend Analysis UwcK 2016 2017 2018 2019 2021 2022 2023 †gvU †iv‡ai Ici ZvcgvÎvi cÖfve Ñ Ñ Ñ 1 1 2 – 4 †iva I Av‡cwÿK †iva Ñ 1 Ñ Ñ Ñ 6 – 7 Ry‡ji Zvcxq wμqvi m~Î, Zwor ÿgZv, Zwor kw3 Ñ 1 Ñ 3 6 8 5 23 †Kv‡li Af ̈šÍixY †iva Ges Zwo”PvjK ej Ñ Ñ Ñ Ñ 2 Ñ – 2 †iv‡ai †kÖwY I mgvšÍivj mgš^q ms‡hvM, Zzj ̈ †iva 2 6 1 3 1 5 – 18 eZ©bxi cÖevngvÎv 2 6 1 2 9 4 4 28 †Kv‡li †kÖwY I mgvšÍivj ms‡hvM 1 Ñ 1 2 1 Ñ – 5 In‡gi m~Îvbymv‡i †fv‡ëR Kv‡i›U †jLwPÎ Ñ 1 Ñ Ñ Ñ 1 1 3 wKk©‡di m~Î: m~‡Îi aviYv, eZ©bx‡Z e ̈envi 1 1 1 1 2 Ñ 4 10 ûBU‡÷vb eaxR bxwZ Ñ 1 Ñ 1 2 4 2 10 A ̈vwgUvi I †fvëwgUv‡ii cvV Ñ Ñ Ñ Ñ 3 1 – 4 A ̈wgUvi Gi cvjøv e„w× I kv‡›Ui e ̈envi 1 1 Ñ Ñ 3 6 1 12 * we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb| weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| DÏxcKwU jÿ ̈ Ki: [Xv. †ev. 23] 8 B 16 32 D A C P Q R S 24 G r = 0.5 E = 12V (K) X = 16 (K) 1 Kzj¤^ Pv‡R©i msÁv `vI| (L) eZ©bx‡Z †Kv‡li ZworPvjK ej m¤ú~Y© Kvh©Ki nq bv †Kb?Ñ e ̈vL ̈v Ki| (M) M ̈vjfv‡bvwgUvi wew”Qbœ Ae ̄’vq Zwor cÖevn wbY©q Ki| mgvavb: 8  16  24  32  12 V 0.5  Req = (24–1 + 56–1 ) –1  Req = 16.8   Zwor cÖevn, I = V Req + r  I = 12 16.8 + 0.5  I = 0.6936 A  M ̈vjfv‡bvwgUvi wewQbœ Ae ̄’vq Zwor cÖevn 0.6936 A| (Ans.) (N) DÏxc‡Ki ‘X’ †ivawU, wea‡R cÖ`Ë †Kv‡bv GKwU †iv‡ai mv‡_ e ̈envi K‡i mvg ̈e ̄’v m„wó m¤¢e wKbv? MvwYwZK e ̈vL ̈v `vI| mgvavb: eZ©bx‡Z S †iva‡K cwieZ©b K‡i mvg ̈ve ̄’v m„wó Kiv m¤¢e| Avgiv Rvwb, mvg ̈ve ̄’vq, P Q = R S  S = 16  24 8 = 48   S †iv‡ai 32  Gi mv‡_ DÏxc‡Ki 16  †iva †kÖwY‡Z mshy3 Ki‡j, S = 32 + 16 = 48  n‡e| (Ans.) 2| R=20 S M T (K) N E = 12V evwZ [iv. †ev. 23]

Pj Zwor  HSC Prep Papers 3 mgvavb: I 1  G 90  3A Current divider low Abyhvqx, I = 90 90 + 1  3 = 2.96 A (Ans.) (N) DÏxcK Abymv‡i Kx e ̈e ̄’v MÖnY Ki‡j 30 A ZworcÖevn gvcv m¤¢e? mgvavb: Avgiv Rvwb, mv‡›Ui †iva, S = G n – 1 = 30 30 3 – 1 = 10   S = R + 1  10 = R + 1  R = 9  A_©vr, 1  †iv‡ai mv‡_ 9  †iva †kÖwY‡Z hy3 Ki‡j M ̈vjfv‡bvwgUvi 30 A Zwor cÖevn gvc‡Z mÿg n‡e| (Ans.) 5| R1 = 40 R2 = 4 R3 = 16 E = 9 V r = 1  R4 = 15 R5 = 10 S [P. †ev. 23] (K) wgUvi eaxR Kx? (L) †Kv‡bv evwZi Mv‡q ‘100 W – 220 V’ †jLv Av‡Q| ej‡Z Kx eyS? (M) eZ©bxi Pvwe S †Lvjv Ae ̄’vq g~j cÖevngvÎv KZ? mgvavb: 9 V 1  40  4  Pvwe S †Lvjv Ae ̄’vq, Zzj ̈ †iva, Req = 40 + 4 = 44   g~j Zwor cÖevn, I = V Req + r  I = 9 44 + 1  I = 0.2 A (Ans.) (N) DÏxc‡Ki eZ©bxi Pvwe S eÜ Ae ̄’vq 10 min Zwor cÖevn Pvjbvi d‡j R4 †iv‡ai wdDRwU M‡j hv‡e wK? †Zvgvi DËi MvwYwZK we‡køl‡Y `vI| [wdDRwU 300 J Zv‡c M‡j hv‡e] mgvavb: I 40  4  15  16  10  I2 I1 I 40  4  I1 16  RP1 44  22  I1 I RP1 = 15  10 15 + 10 = 6   Req = 44  22 44 + 22 = 14.67   g~j Zwor cÖevn, I = V Req + r  I = 9 14.67 + 1  I = 0.5743 A Current divider law Abyhvqx, I1 = 44 44 + 22  I = 0.3828 A Avevi, I2 = 10 10 +15  I1 = 0.1531 A  Drcbœ Zvc, H = I2 2 Rt = 0.15312  15  (10  60)  H = 210.95 J < 300 J myZivs †iv‡ai wdDRwU M‡j hv‡e bv| (Ans.) 6| 12.5  †iv‡ai GKwU Zvi‡K †U‡b wØ ̧Y K‡i GKwU wnUv‡ii KzÐjx •Zwi K‡i G‡Z 220 V mieivn jvB‡b hy3 Kiv n‡jv| KzÐjx‡K 30C Gi cvwb‡Z wbgw3⁄4Z K‡i 305 sec. we`y ̈r Pvjbv Kiv n‡jv| a‡i wb‡Z n‡e m¤ú~Y© Zwor kw3 Zvc kw3‡Z iƒcvšÍwiZ n‡q‡Q| [P. †ev. 23] (K) Kvk©‡di 2q m~Î wee„Z Ki| (L) kv‡›Ui †iva k~b ̈ ev Amxg †KvbwUI bqÑ e ̈vL ̈v Ki|