Question

The peak and rms value of current in AC circuit. The current is represent by the equation $i=5\sin \left( 300t-\dfrac{\pi }{4} \right)$where t is in seconds and i is in ampere.
A. 5A, 3.535A
B. 5A, 5.53A
C. 3A, 3.53A
D. 6.25A, 5.33A

Answer 1

Hint: As a very first step, one could read the question well and hence write down the points that seem important. Then you could recall the standard expression in which the current in an AC circuit is expressed. Compare it with the given expression and note down the values. Now recall the expression for rms current and find the answer.

Formula used:
Current in an AC circuit,
$i={{i}_{0}}\sin \left( \omega t+\phi \right)$

Complete step-by-step solution:
In the question, we are given the expression that represents the current in an AC circuit which is given by,
$i=5\sin \left( 300t-\dfrac{\pi }{4} \right)$
Here, t is given in seconds and I in amperes which are both known to be their respective SI units. We are supposed to find the peak and rms values of the current from the given information.
Let us recall the standard expression for the current,
$i={{i}_{0}}\sin \left( \omega t+\phi \right)$
Here, we could conclude that the peak value of the current is
${{i}_{0}}=5A$…………………………………. (1)
The angular frequency of the AC current source would be,
$\omega =300{{s}^{-1}}$
The phase constant would be,
$\phi =\dfrac{\pi }{4}$
Now, let us recall the expression for rms current,
${{i}_{rms}}=\dfrac{{{i}_{0}}}{\sqrt{2}}$
Substituting (1) we get,
${{i}_{rms}}=\dfrac{5}{\sqrt{2}}=3.535A$…………………………………… (2)
From (1) and (2) we found the peak and rms values of the current to be 5A and 3.535A respectively.
Hence, option A is correct.

Note: Very similar to the given expression for current in an AC circuit, we have the expression for voltage in an AC circuit given by,
$V={{V}_{0}}\sin \left( \omega t-\phi \right)$
Here, ${{v}_{0}}$is known to be the peak voltage and we also have rms value for voltage given by a very similar expression as,
$V=\dfrac{{{V}_{0}}}{\sqrt{2}}$