Question

Three sources of emf ${V_0}\sin \omega t$ ,${V_0}\cos \omega t$ , and ${V_0}$ are connected in series as shown. RMS value of current in the circuit is?

Answer 1

Here we have to use the concept of RMS values of AC and DC signals and finding out their Net value along with the use of Ohm's Law.
Formula used:
From ohms law the current $I = \dfrac{V}{R}$
So, ${I_{rms}} = \dfrac{{{V_{rms}}}}{R}$
For a sinusoidal AC source as ${V_{rms}} = \dfrac{{{V_0}}}{{\sqrt 2 }}$
For the three different voltage sources the resultant of all these SRM voltages,
${V_{netRMS}} = \sqrt {{{({V_{1rms}})}^2} + {{({V_{2rms}})}^2} + {{({V_{3rms}})}^2}}$

Complete step by step answer:
Here we have to find the RMS value of current,
What we have here is some voltage sources and resistance. ohms will give us a relation connecting current and voltage.
So from ohm's law, we can calculate current as,$I = \dfrac{V}{R}$
The term RMS stands for Root Mean Squared. The RMS value is the square root of the mean (average) value of the squared function of the instantaneous values. The symbols used for defining an RMS value are ${V_{rms}},{I_{rms}}$ .
So, ${I_{rms}} = \dfrac{{{V_{rms}}}}{R}$
To find ${V_{rms}}$ , we have to consider each source and have to find Net ${V_{rms}}$ in the circuit.
The sources ${V_0}\sin \omega t$ and ${V_0}\cos \omega t$ are sinusoidal AC sources with peaks at ${V_0}$ . we know the value of ${V_{rms}}$ for a sinusoidal AC source as ${V_{rms}} = \dfrac{{{V_0}}}{{\sqrt 2 }}$
For DC source ${V_{rms}} = {V_{dc}}$
Since, we have, so we three different voltage sources here have to find resultant of all these SRM voltages,
${V_{netRMS}} = \sqrt {{{({V_{1rms}})}^2} + {{({V_{2rms}})}^2} + {{({V_{3rms}})}^2}}$
${V_{1rms}} = {V_{dc}} = {V_0}$
${V_{2rms}} = {V_{3rms}} = {V_{ac}} = \dfrac{{{V_0}}}{{\sqrt 2 }}$
Now substituting all these values in the Net RMS value of Voltage, we get
${V_{rms}} = \sqrt {{{({V_0})}^2} + {{(\dfrac{{{V_0}}}{{\sqrt 2 }})}^2} + (\dfrac{{{V_0}}}{{\sqrt 2 }}} {)^2}$
Simplifying this equation,
${V_{rms}} = \sqrt {{V_0}^2 + \dfrac{{{V_0}^2}}{2} + \dfrac{{{V_0}^2}}{2}}$
$= \sqrt {2{V_0}^2}$
$= \sqrt 2 {V_0}$
Now we got the value of RMS voltage, so from this by using ohm's law we can find out ${I_{rms}}$
${I_{rms}} = \dfrac{{{V_{rms}}}}{R} = \dfrac{{\sqrt 2 {V_0}}}{R}$
So the answer is $\dfrac{{{V_0}\sqrt 2 }}{R}$.

Note:
If you have given some sinusoidal waveform, with the known value of its peak, then we can use these relations. Other than peak values, sometimes it may give average values or most probable values. That time, the equations will have some modifications.