An alternating emf E = 440 sin 100 πt is applied to a circuit containing an inductance of √2πH. If an a.c. ammeter is connected in the circuit, its reading will be:
Question
An alternating emf E = 440 sin 100 πt is applied to a circuit containing an inductance of √2πH. If an a.c. ammeter is connected in the circuit, its reading will be:
Solution
To solve this problem, we need to use the formula for the current in an inductive circuit, which is given by I = E / XL, where E is the emf and XL is the inductive reactance.
The inductive reactance XL is given by the formula XL = 2πfL, where f is the frequency and L is the inductance.
From the given emf E = 440 sin 100πt, we can see that the coefficient of t is 100π, which is equal to 2πf. Therefore, the frequency f is 50 Hz.
Now, we can calculate the inductive reactance XL:
XL = 2πfL = 2π * 50 Hz * √2π H = 100π^2 Hz.
Now, we can calculate the current I:
I = E / XL = 440 V / 100π^2 Hz = 0.44/π A.
Therefore, the reading on the AC ammeter will be 0.44/π A.
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