Step 1: Identify Current Components
The given current consists of:
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
Common Mistakes to Avoid
Conclusion
The correct RMS value of the current is 3 (10 Amp).
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: \[ i \geq 10 \, \text{Amp} \] \[ \boxed{i = 10 \, \text{Amp}} \]

Let \( i_C, i_L, \) and \( i_R \) be the currents flowing through the capacitor, inductor, and resistor, respectively, in the circuit given below. The AC admittances are given in Siemens (S).
Which one of the following is TRUE?

A simplified small-signal equivalent circuit of a BJT-based amplifier is given below.
The small-signal voltage gain \( \frac{V_o}{V_S} \) (in V/V) is _________.

Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.