Calculating Voltage and Current for an AC Source | 180sin377t Equation Explained

[Dr3w529]

The voltage of an AC source is given by the equation v=180sin377t where v is the instantaneous voltage and t is in seconds. (a) what is the rms value of the voltage? (b) what is the frequency of the ac source? (c) if this voltage is connected to a 40 ohm resistor, find the equation that describles the current I as a function of time.

My Work:
Vrms= Vpeak/(sq rt)2
180sin377t/(sqrt)2
=1.6t/(sqrt)2
=1.13t
I=V/z
1.6t/(sqrt)2/40
=.028t

 

[Curious3141]

The voltage of an AC source is given by the equation v=180sin377t where v is the instantaneous voltage and t is in seconds. (a) what is the rms value of the voltage? (b) what is the frequency of the ac source? (c) if this voltage is connected to a 40 ohm resistor, find the equation that describles the current I as a function of time.

My Work:
Vrms= Vpeak/(sq rt)2
180sin377t/(sqrt)2
=1.6t/(sqrt)2
=1.13t
I=V/z
1.6t/(sqrt)2/40
=.028t

The voltage function is properly represented by V = 180\sin({377t})

A few things to note.

1) A general voltage function takes the form V = V_0\sin(\omega t) where V_0, \omega respectively represent the peak voltage and the angular frequency respectively.

2) The angle \omega t is generally assumed to be in radians, not degrees.

In calculating the rms voltage, you have the right formula V_{rms} = \frac{V_0}{\sqrt{2}} But you used the wrong value for the peak voltage. You shouldn't touch whatever comes after the sin, because that's the angular (phase) component. Just use the 180 as the peak voltage and recalculate.

For the frequency, look up the relationship between angular frequency and frequency. Remember that we're working in radians, so you should expect to see "pi" cropping up. The angular frequency here is given as 377 radians/second.

For the last part, remember that a resistor is a purely resistive load and the current will be in phase with the voltage. The voltage and the current will be related by Ohm's law. The current will change with time, matching the shape of the voltage waveform. Knowing Ohm's law, it's a simple matter to figure out the equation of the current as a function of time.