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
Solution and Explanation
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).
Top Questions on AC Circuits
For the AC circuit shown in the figure, $ R = 100 \, \text{k}\Omega $ and $ C = 100 \, \text{pF} $, and the phase difference between $ V_{\text{in}} $ and $ (V_B - V_A) $ is 90°. The input signal frequency is $ 10^x $ rad/sec, where $ x $ is:
- JEE Main - 2025
- Physics
- AC Circuits
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 \( i_C, i_L, \) 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 3-input majority logic gate has inputs \( X \), \( Y \), and \( Z \). The output \( F \) of the gate is logic ‘1’ if two or more of the inputs are logic ‘1’. The output \( F \) is logic ‘0’ if two or more of the inputs are logic ‘0’.
Which one of the following options is a Boolean expression of the output \( F \)? - The ideal BJT in the circuit given below is biased in the active region with a \( \beta \) of 100. If \( I_B \) is 10 µA, then \( V_{CE} \) (in Volts, rounded off to two decimal places) is _________.
Questions Asked in JEE Main exam
If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:
- JEE Main - 2025
- Trigonometric Identities
If the area of the region $\{ (x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to
- JEE Main - 2025
- Area between Two Curves
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
- JEE Main - 2025
- Matrices
Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
- JEE Main - 2025
- Determinants
Let $C$ be the circle $x^2 + (y - 1)^2 = 2$, $E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on the $x$-axis and $y$-axis respectively. Let the straight line $x + y = 3$ touch the curves $C$, $E_1$, and $E_2$ at $P(x_1, y_1)$, $Q(x_2, y_2)$, and $R(x_3, y_3)$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$, the value of $9(x_1 y_1 + x_2 y_2 + x_3 y_3)$ is equal to
- JEE Main - 2025
- Coordinate Geometry