An AC current is represented as:
$
i = 5\sqrt{2} + 10 \cos\left(650\pi t + \frac{\pi}{6}\right) \text{ Amp}
$
The RMS value of the current is:
Show Hint
- 50 Amp
- 100 Amp
- 10 Amp
- \( 5\sqrt{2} \) Amp
The Correct Option is C
Approach Solution - 1
Step 1: Identify Current Components
The given current consists of:
- A DC component: \( I_{\text{DC}} = 5\sqrt{2} \) Amp (constant term)
- An AC component: \( i_{\text{AC}}(t) = 10 \cos\left(650\pi t + \frac{\pi}{6}\right) \) Amp
Step 2: RMS Value of AC Component
For any sinusoidal current \( I_{\text{peak}} \cos(\omega t + \phi) \), the RMS value is: \[ I_{\text{AC,rms}} = \frac{I_{\text{peak}}}{\sqrt{2}} \] Here, \( I_{\text{peak}} = 10 \) Amp, so: \[ I_{\text{AC,rms}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \text{ Amp} \]
Step 3: Total RMS Value Calculation
When both DC and AC components are present: \[ I_{\text{rms}} = \sqrt{I_{\text{DC}}^2 + I_{\text{AC,rms}}^2} \] Substituting the values: \[ I_{\text{rms}} = \sqrt{(5\sqrt{2})^2 + (5\sqrt{2})^2} \] \[ I_{\text{rms}} = \sqrt{50 + 50} \] \[ I_{\text{rms}} = \sqrt{100} \] \[ I_{\text{rms}} = 10 \text{ Amp} \]
Verification
- \( (5\sqrt{2})^2 = 25 \times 2 = 50 \)
- Sum of squares: \( 50 + 50 = 100 \)
- \( \sqrt{100} = 10 \)
Common Mistakes to Avoid
- Ignoring the DC component's contribution
- Forgetting to divide the peak value by \( \sqrt{2} \) for AC RMS
- Adding RMS values directly instead of their squares
Conclusion
The correct RMS value of the current is 3 (10 Amp).
Approach Solution -2
Given: \[ i = 5\sqrt{2} + 10 \cos\left(650\pi t + \frac{\pi}{6}\right) \] Squaring both sides: \[ i^2 = 50 + 100 \cos^2\left(650\pi t + \frac{\pi}{6}\right) \] Expanding: \[ i^2 = 50 + 100 \cos^2\left(650\pi t + \frac{\pi}{6}\right) + (2)(5\sqrt{2})(10) \cos\left(650\pi t + \frac{\pi}{6}\right) \] Simplifying: \[ i^2 = 50 + \frac{100}{2} + 0 \] \[ i^2 \geq 100 \] Taking the square root: \[ i \geq 10 \, \text{Amp} \] \[ \boxed{i = 10 \, \text{Amp}} \]
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