CODE - 607383-2017 Fall, PHYS 212 (cook) Test2 17-Y01 5 pt (c18p10) A piece of N

Question

CODE - 607383-2017 Fall, PHYS 212 (cook) Test2 17-Y01 5 pt (c18p10) A piece of Nic A when a potential of 1.5 resistance of the wire? (ohms Name: (in ohms) 5 AO 1.18 BO 1.7 EO 5.2 HO 1.5 5 pt A capacitor consists of two closely spaced metal con- uctors of large area, separated by a thin insulating foil. It has an electrical capacity of 3600.0 F and is charged to a potential difference of 69.0 V. Calculate the amount of energy stored in the capacitor. DO 3.59 GO 1.09×10' A 130W street lamp is opera much energy does it take to o in J) 1. AO 4.08 BO 5.91 CO 8.57 DO 1.24 x 10 EO 1.80 x 10 FO 2.61 x 105 pt Express your answer in (in kweh) 6. AO 1.07 x 102 GO 3.79 × 101 HO 549 × 101 5 pt (8c25p45) A certain substance has a dielectric constant of 2.8 and a dielectric strength of 18.0 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what BO 1.4: EO 33. HO 7.9. DO 2.52 × 102 GO 5.94 × 102 tain a capacitance of 7.00 × 10-8 F and to ensure that the capacitor will be able to withstand a potential difference of 15 ptIDue never 5.4 kV? (in m2) 2, AO 1.78 × 10-1 BO 2.22 x 10-1 CO 2.78 × 10-1 (c18p16) Calculate the resist nichrome wire 2.50m long. T gauge wire is 0.5176mm2 DO 3.47 x 10-1 EO 4.34 × 10-1 FO 5.42 × 10-1 GO 6.78 × 10-1 HO 8.47 × 10-1 Resistivity for ('calculator" no aluminium: 2.65 e-8 copper: 1.72 e-8 gold: 2.24 e-8 iron: 9.71 e-8 rome: 1.00 e-6 5 pt (c26p80_66) Three capacitors are connected in parallel. nich has plate area A = 2.00 × 10-2 m: and plate spacing Phatinum: silver: 1.59 e-8 d = 2.60 × 10-3 m. what must be the spacing of a single capacitor of plate area A if its capacitance equals that of the uge parallel combination? (in a) 3, AO 284 x 10-4 BO 3.55 x 0-4 CO 4.44 × 10-4 (in ohms) 7. AO 7.54×10-1 BO 100 DO 5.55 x 10-4 EO 6.93 × 10-4 FO 8.67 × 10-4 DO 2.30 EO 3.33

Explanation / Answer

(a) C = 3600 uF = 3600 x 10-6 F V = 69 volt

so the energy stored in the capacitor

E = CV2 / 2

= 0.5 x 3600 x 10-6 x 69 x 69 = 8.56 J so option C will correct Ans

(b) dielectric constant K = 2.8 dielectric strength E = 18 MV/m = 18 x 106 m

as we know that E = V/d

V = 5.4 KV = 5400 volt so d = 5400 / (18 x 106) = 0.0003 m

and C = 7 x 10-8 F

so as we know that C = KeoA / d

(7 x 10-8) = (2.8 x 8.85 x 10-12) A / 0.0003

A = (0.0021 x 10-8) / (24.78 x 10-12)

= 0.8474 m2 = 8.474 x 10-1 m2   so option H will correct Ans

(c) here A = 2 x 10-2 = 0.02 m2   and d = 2.60 x 10-3 m = 0.00260 m

as we know that C = eoA/d

C1 = (8.85 x 10-12 x 0.02) / 0.00260

C1 = 6.807 x 10-11 F

here plate area and spacing for each capacito are same so

C1 = C2 = C3 = 6.807 x 10-11 F

so the total capacitance of three pallal combination

C = C1 + C2 + C3

C = 20.421 x 10-11 F

now according to question

A = 0.02 m2 and Cn = C = 20.421 x 10-11 F

so Cn = eoA / dn

dn = (8.85 x 10-12 x 0.02) / (20.421 x 10-11)

dn = 0.0008667 m = 8.667 x 10-4 m so option F will correct Ans

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