Question

The output voltage of an ac generator is given by δv = (1.09 ✕ 102 V)sin(31πt). The generator is connected across a 0.500 h inductor. Find the following. (a) frequency of the generator.

Answer 1

The frequency of the generator that is connected across a 0.500 H inductor and has a voltage given by δv = (1.09 × 10² V)sin(31πt) is 15.5 Hz.

Step-by-step explanation:

The frequency of the generator can be determined by examining the angular frequency ω in the given output voltage equation δv = (1.09 × 102 V)sin(31πt). The angular frequency ω is equal to 31π rad/s. To find the frequency f, we can use the relationship between angular frequency and frequency, which is ω = 2πf. Hence, the frequency f is ω/(2π) = (31π rad/s) / (2π) = 15.5 Hz.