Answer :
Final answer:
The equation for the instantaneous current through a circuit with an alternating e.m.f. applied to an inductance is I = 2.2 sin 100πt A. An a.c. galvanometer in the circuit would display the RMS value of the current as approximately 1.555 A.
Explanation:
The application of the alternating e.m.f. E = 220 sin 100πt to a circuit with an inductance of (1/π) henry leads us to the procedure of calculating the instantaneous current through the circuit. The inductive reactance (X_L) of the inductor can be calculated with the formula X_L = ωL, where ω is the angular frequency of the applied voltage, and L is the inductance. In this case, ω = 100π rad/s. Therefore, the inductive reactance is ωL = (100π)(1/π) = 100 Ω. The instantaneous current through the circuit can be derived using Ohm's law for AC circuits, which is I = V/Z, where V is the applied voltage and Z is the impedance. Since only inductance is present, Z = X_L. Thus, the equation for the current becomes I = E/X_L = 220 sin 100πt / 100, which simplifies to I = 2.2 sin 100πt A.
Regarding the reading of an a.c. galvanometer connected in the circuit, it would measure the root mean square (RMS) value of the current, not the instantaneous current. The RMS value of an AC current is I_{RMS} = I_{max}/√2, where I_{max} is the maximum (or peak) value of the current. Therefore, the RMS current in this circuit is 2.2/√2 A, which is approximately 1.555 A. This value represents the effective current that would be displayed on the galvanometer.