If the alternating current as a function of time in an AC heating coil is given by the expression l = 5.6 sin 120 × 3.14t, what is the RMS current and AC frequency?

To find the RMS current, we need to first find the peak current. The peak current is the maximum value of the current, which is the coefficient of the sine function. In this case, the peak current is 5.6 A.

The RMS current is given by the formula:

Irms = Ipeak / √2

Substitute the peak current value into the formula:

Irms = 5.6 / √2 ≈ 3.96 A

Therefore, the RMS current is approximately 3.96 A.

To find the AC frequency, we look at the coefficient of t in the sine function, which is 120 × 3.14. The frequency of an AC signal is given by the coefficient of t multiplied by 2π. Therefore, the AC frequency is:

f = 120 × 3.14 × 2π ≈ 753.98 Hz

Therefore, the AC frequency is approximately 753.98 Hz.

The RMS current is:

$$I_{rms}=\frac{I_m}{\sqrt{2}}=\frac{5.6}{\sqrt{2}}=3.96\text{ A}$$

AC frequency is:

$$f=\frac{1}{T}=\frac{\omega}{2\pi}=\frac{120\times3.14}{2\pi}=\text{60 Hz}$$