Title 3. P2C3 Current Electricity_With Solve.pdf ✅

Pj Zwor  Mastery Practice Sheet 3 = 0.0485 W (Ans.) 13| we`y ̈‡Zi e ̈envi Kgv‡bvi j‡ÿ ̈ GKwU mvaviY 60 W GSL evwZ‡K GKwU 13 W CFL evwZ w`‡q e`jv‡bv nj| evj¦ `ywUi g~j h_vμ‡g Tk. 30 Ges Tk. 250 cÖwZ BDwbU we`y ̈‡Zi `vg Tk. 4 n‡j GK eQ‡ii g‡a ̈ evj¦wU e`jv‡bvi LiP DVv‡Z cÖwZw`b M‡o KZ NÈv CFL evj¦wU‡K R¡vjv‡Z n‡e? [BUET 12-13] mgvavb: (250 – 30) = (60 – 13) × 10–3 × t × 365 × 4  t = 220 4 × 0.047 × 365 hr = 3.206 hr (Ans.) 14| 6 V-Gi GKwU e ̈vUvixi Af ̈šÍixY †iva 0.25 | Ab ̈ GKwU 0.5  Af ̈šÍixY †ivawewkó 3 V e ̈vUvixi mv‡_ mgvšÍiv‡j ms‡hvM Ki‡j D3 mgev‡qi cÖvšÍ؇qi wefe cv_©K ̈ wbY©q Ki| [BUET 11-12] mgvavb: I I A C B 3 V 6 V 0.25  0.5  D V ABCDA jy‡c KVL e ̈envi K‡i cvB, – 6 + 3 + 0.5I + 0.25I = 0  0.75I = 3 I = 4 A  VAC = 3 + 0.5I = (3 + 2) V = 5 V (Ans.) 15| GKwU •e`y ̈wZK B ̄¿x‡Z ‘220 V – 1000 W’ †jLv Av‡Q| Bw ̄¿wU 200 V jvB‡b hy3 n‡q 2 NÈv Pj‡j KZ BDwbU we`y ̈r kw3 LiP Ki‡e? [BUET 09-10] mgvavb: R = V 2 P = 2202 1000  = 48.4  W = V 1 2 R  t = 2002 48.4 × 10–3 × 2KW-h = 1.653 Unit (Ans.) 16| 5 ohms †ivawewkó GKwU Zvi‡K †U‡b wZb ̧Y j¤^v Kiv nj| j¤^vK...Z ZviwUi †iva wbY©q Ki| [BUET 08-09, 04-05, 02-03] mgvavb: †U‡b j¤^v Ki‡j AvqZb aaæe _vK‡e|  A1L1 = A2.3 L1  A1 = 3 A2  R2 R1 = L2 L1 × A1 A2 = 3 × 3 = 9  R2 = 9 × 5  = 45  (Ans.) 17| 27 C ZvcgvÎvq 1 kW GKwU B‡jKwUaK †KZwj‡Z 2 litre cvwb Av‡Q| †KZwjwU‡K 10 wgwb‡Ui Rb ̈ myBP Ab Kiv n‡jv| hw` Pvicv‡k Zvc n«v‡mi nvi 160 Js–1 nq Z‡e 10 wgwb‡U †KZwji ZvcgvÎv KZ n‡e? [BUET 06-07] mgvavb: Pt = ms  (1000 – 160) × 10 × 60 = 2 × 4200 × ( – 27)   = 87C (Ans.) 18| †kÖwY mgev‡q mw3⁄4Z `ywU cwievnxi †iva 40-ohm hv mgvšÍivj mgev‡q 7.5-ohm nq| cÖwZwU cwievnxi †iva †ei Ki| [BUET 05-06] mgvavb: R1 + R2 = 40 .............. (i) R1R2 R1 + R2 = 7.5  R1R2 = 300 ........... (ii)  R1 – R2 = 20 ........... (iii)  R1 = 30  (Ans.) R2 = 10  (Ans.) 19| GKwU ûBU‡÷vb wea‡Ri cÖ_g Ges wØZxq evû‡Z h_vμ‡g 10  Ges 12  Gi †iva hy3 Av‡Q| hLb PZz_© evû‡Z 20  Gi `ywU †iva mgvšÍivj ms‡hv‡M hy3 nq ZLb weaRwU mvg ̈ve ̄’vq _v‡K| ARvbv †iv‡ai gvb KZ? [BUET 04-05] mgvavb: P Q = R S  10 12 = R 20 || 20  R = 10 × 10 12  = 8.33  (Ans.) 20| GKwU †cv‡UbwmIwgUvi Zv‡i we`y ̈r cÖevn wbqš¿Y K‡i †Kvb we`y ̈r †Kv‡li Rb ̈ 6 m `~‡i wb ̄ú›` we›`y cvIqv †Mj| †KvlwUi `y-cÖv‡šÍi mv‡_ 3 In‡gi GKwU †iva hy3 Ki‡j 4 m `~‡i wb ̄ú›` we›`y cvIqv hvq| †KvlwUi Af ̈šÍixY †iva wbY©q Ki| [BUET 02-03] mgvavb: r =     l2 l1 – 1 R =     6 4 – 1 × 3 = 1.5  (Ans.) 21| 30  Af ̈šÍixY †iv‡ai GKwU M ̈vjfv‡bvwgUvi 500 A Zwor cÖev‡n c~Y© † ̄‹j we‡ÿc †`q| GB M ̈vjfv‡bvwgUvi‡K 2 mA cÖevngvÎv cwigv‡ci Dc‡hvMx Ki‡Z KZ gv‡bi kv›U e ̈envi Ki‡Z n‡e? [BUET 01-02]