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State the principle on which transformer works. Explain its working with construction. Derivean expressionfor ratio of e.m.f.s and currentsin terms ofnumber ofturnsin primaryand secondarycoil. A conductorof any shape, havingarea 40 cm^(2)placed inair is uniformaly chargedwith a charge0.2 muC. Determine the electricintensityat a pointjustoutsideits surface. Also,find mechanicalforceper unitarea of thechargedconductor. [epsi_(0) = 8.85 xx 10^(-12) S.I. units] |
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Answer» Solution :Numerical : Given : ` Q = 0.2 muC= 0.2 xx 10^(-6)C, A = 40 cm^(2) =40 xx 10^(-4) m^(2)` `epsi_(0) = 8.85 xx 10^(-12)` SI units The electricfieldintensity just outsidethe surfaceof a chargedconductorof any shapeis `E = (sigma)/(epsi_(0)) = (Q)/(Aepsi_(0)) [ :' sigma = (Q)/(A)]` `:. E = (0.2xx 10^(-6))/(40 xx 10^(-4) xx 8.85xx 10^(-12))` `:. E = 5.65 xx 10^(6) N/C` Now, the mechanicalforce per unit area of a conductor is `F = 1/2 epsi_(0) E^(2)` `= 1/2 xx 8.85 xx 10^(-12) xx (5.65 xx 10^(6))^(2)` `:. F = 141.25 N//m^(2)` |
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