Question

The root mean square R.M.S value of current $i = 3 + 4\sin \left( {\omega t + \dfrac{\pi }{3}} \right)$ is:
$
  (a){\text{ 5A}} \\
  {\text{(b) }}\sqrt {17} A \\
  (c){\text{ }}\dfrac{5}{{\sqrt 2 }}A \\
  (d){\text{ }}\dfrac{7}{{\sqrt 2 }}A \\
 $

Answer 1

Hint: In this question use the concept that for any equation of the form $A = x + y\sin \left( {\omega t + \dfrac{\pi }{3}} \right)$, the R.M.S value of A can be calculated as ${A_{r.m.s}} = \sqrt {{x^2} + {{\left( {\dfrac{y}{{\sqrt 2 }}} \right)}^2}} $. Compare the given equation of current with the standard equation to get the R.M.S value.

Complete Step-by-Step solution:
Given value of current is
$i = 3 + 4\sin \left( {\omega t + \dfrac{\pi }{3}} \right)$
Let $A = x + y\sin \left( {\omega t + \dfrac{\pi }{3}} \right)$
The R.M.S value of A is calculated as
${A_{r.m.s}} = \sqrt {{x^2} + {{\left( {\dfrac{y}{{\sqrt 2 }}} \right)}^2}} $So use this property in given equation we have,
$ \Rightarrow {i_{r.m.s}} = \sqrt {{3^2} + {{\left( {\dfrac{4}{{\sqrt 2 }}} \right)}^2}} $
Now simplify the above equation we have,

$ \Rightarrow {i_{r.m.s}} = \sqrt {9 + \left( {\dfrac{{16}}{2}} \right)} = \sqrt {9 + 8} = \sqrt {17} $ Amp.
So this is the required answer.
Hence option (B) is the correct answer.

Note – In general mathematics R.M.S value is also known as quadratic mean. The R.M.S value of a set of values is the square root of arithmetic mean of the squares of the values or the square of the function that defines the continuous waveform. For general case of set of n values $\left\{ {{x_1},{x_2},{x_3},............,{x_n}} \right\}$ the R.M.S is ${x_{RMS}} = \sqrt {\dfrac{1}{n}\left( {{x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2....... + {x_n}^2} \right)} $.