An alternating voltage given by $V = 140(\sin 314t)$ is connected across a pure resistor of $50\Omega $. Find
(A) the frequency of the source
(B) the rms current through the resistor

Hint: In this question, we can compare the given equation with the standard alternating voltage equation and get the required values to find the frequency and maximum current. Then use the relation between maximum current and its rms value to get the answer.
Formula used:
$\omega = 2\pi f$
${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$

Complete step by step answer We know that the alternating voltage is given by the equation
$V = {V_0}\sin \omega t$where ${V_0}$is the peak voltage and $\omega $is the angular frequency
So, on comparing above equation to the given equation, we get
${V_0} = 140$and $\omega = 314$
To find frequency, we know the relation between angular frequency and time period that is $\omega = \dfrac{{2\pi }}{T} = 2\pi f \Rightarrow f = \dfrac{\omega }{{2\pi }}$where f is the frequency
Putting the value,
$f = \dfrac{{314}}{{2 \times 3.14}} = 50Hz$
Answer (i) The frequency of source is 50Hz
The rms or root mean square current is the square root of the average square of instantaneous current for a full cycle.
We also know that rms current is 0.707 of the peak value
$ \Rightarrow {I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$
Let us first find the value of peak current , ${I_0} = \dfrac{{{V_0}}}{R}$where ${V_0}$is the peak voltage and R is the resistance
On putting the values in the equation, we get ${I_0} = \dfrac{{140}}{{50}} = \dfrac{{14}}{5}$
So, the rms value of current is ${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }} = \dfrac{{14}}{{5 \times 1.41}} \approx 2A$

Answer (ii)The rms current is 2A

Note:
Devices like ammeter and voltmeter measure the current or voltage and give its rms value. Rms value is also called ‘virtual value’ or ‘effective value’ and has great significance.
The alternating current supplied in our home is the rms value. When we say 230V is supplied it means that it is the rms value of voltage supplied. If we calculate the peak voltage ${V_0} = \sqrt 2 \times {V_{rms}} = \sqrt 2 \times 230 = 325V$
This means that voltage supplied varies from 325V to -325V. This is the reason it is said that a 230V a.c. is more deadly than 230V d.c.