The inductive reactance is:
\[ X_L = \omega L = 100 \times 1 = 100 \, \Omega. \]
The capacitive reactance is:
\[ X_C = \frac{1}{\omega C} = \frac{1}{100 \times 20 \times 10^{-6}} = 500 \, \Omega. \]
The total impedance is:
\[ Z = \sqrt{(X_L - X_C)^2 + R^2} = \sqrt{(100 - 500)^2 + 300^2}. \]
Simplifying:
\[ Z = \sqrt{(-400)^2 + 300^2} = \sqrt{160000 + 90000} = 500 \, \Omega. \]
The rms current is:
\[ i_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{50}{500} = 0.1 \, \text{A}. \]
The rms voltage across the capacitor is:
\[ V_{\text{rms, capacitor}} = X_C \cdot i_{\text{rms}} = 500 \times 0.1 = 50 \, \text{V}. \]
Find output voltage in the given circuit.
A | B | Y |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |
The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance covered by the object between \( t = 0 \) to \( t = 4s \)?
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $