An alternating current is represented by the equation, $\mathrm{i}=100 \sqrt{2} \sin (100 \pi \mathrm{t})$ ampere. The RMS value of current and the frequency of the given alternating current are
An alternating current is represented by the equation, $\mathrm{i}=100 \sqrt{2} \sin (100 \pi \mathrm{t})$ ampere. The RMS value of current and the frequency of the given alternating current are
Show Hint
- $100 \sqrt{2} \mathrm{~A}, 100 \mathrm{~Hz}$
- $\frac{100}{\sqrt{2}} \mathrm{~A}, 100 \mathrm{~Hz}$
- $100 \mathrm{~A}, 50 \mathrm{~Hz}$
- $50 \sqrt{2} \mathrm{~A}, 50 \mathrm{~Hz}$
The Correct Option is C
Approach Solution - 1
2. Frequency of the current: \[ f = \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} = 50 \mathrm{~Hz} \] Therefore, the correct answer is (3) $100 \mathrm{~A}, 50 \mathrm{~Hz}$.
Approach Solution -2
We are given the alternating current equation:
\[ i = 100\sqrt{2} \sin(100\pi t) \]We need to find the RMS value of the current and the frequency of the alternating current.
Concept Used:
The general equation of an alternating current is given by:
\[ i = I_0 \sin(\omega t) \]where \( I_0 \) is the peak current (maximum current) and \( \omega = 2\pi f \) is the angular frequency. The RMS value of the current is:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]Step-by-Step Solution:
Step 1: Identify the peak current \( I_0 \) from the given equation.
\[ I_0 = 100\sqrt{2} \, \text{A} \]Step 2: Calculate the RMS value of the current using \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \).
\[ I_{\text{rms}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100 \, \text{A} \]Step 3: Determine the angular frequency \( \omega \) from the given equation.
\[ \omega = 100\pi \]Step 4: Relate angular frequency to linear frequency \( f \) using \( \omega = 2\pi f \).
\[ 100\pi = 2\pi f \] \[ f = 50 \, \text{Hz} \]Final Computation & Result:
The RMS current is \( 100 \, \text{A} \) and the frequency is \( 50 \, \text{Hz} \).
Final Answer: \( I_{\text{rms}} = 100 \, \text{A}, \, f = 50 \, \text{Hz} \)
Correct Option: (3) 100 A, 50 Hz
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