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प्रश्न
Alternating emf of e = 220 sin 100 πt is applied to a circuit containing an inductance of (1/π) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?
उत्तर
Data: e = 220 sin 100 πt, L = `(1/pi)`H
Comparing e = 220 sin 100 πt with
e = e0 sin ωt, we get
ω = 100 π
∴ ωL = (100 π)`(1/π)` = 100 Ω
∴ The instantaneous current through the circuit
= i = `"e"_0/(omega"L") sin (100π"t" - π/2)`
`= 220/100 sin (100π"t" - π/2)`
= 2.2 sin (100πt - π/2) in ampere
`"i"_"rms" = "i"_0/sqrt2 = 2.2/1.414 = 1.556 "A"` is the reading of the AC galvanometer connected in the circuit.
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