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प्रश्न
A coil having self-inductance of 0.7 H and resistance of 165 Ω is connected to an a.c. source of
275V,50Hz. If π = `22/7`
Calculate:
(i) Reactance of the coil
(ii) Impedance of the coil ‘
(iii) Current flowing through the coil
उत्तर
l = 0.7 H , R = 165 Ω , V = 275 V
f = 50 Hz , π = `22/7`
(i) XL = 2 π f L [XL = reactance]
`= 2 xx 22/7 xx 50 xx 0.7 = 220 Omega`
(ii) Impedance (Z) of the coil
Z = `sqrt("R"^2 +(2pi "f" "L")^2) = sqrt((165)^2 + (220)^2) = sqrt75625 = 275 Omega`
(iii) Current (I) in the coil
= `"V"/"Z" = (275"V")/(275Omega) = 1"A"`
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