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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.

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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?

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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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पाठ 13: AC Circuits - Exercises [पृष्ठ ३०५]

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बालभारती Physics [English] Standard 12 Maharashtra State Board
पाठ 13 AC Circuits
Exercises | Q 16. | पृष्ठ ३०५

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