Question

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?

Answer 1

Data: e = 220 sin 100 πt, L = (\(\frac{1}{\pi}\) )H 

Comparing e = 220 sin 100 πt with e = e₀ sin ωt, we get

ω = 100 π 

∴ ωL = (100 π) (\(\frac{1}{\pi}\) ) = 100 Ω 

∴ The instantaneous current through the circuit

is the reading of the AC galvanometer connected in the circuit.