For DC voltage:
\[ R = \frac{V}{I} = \frac{100}{5} = 20 \, \Omega \]
For AC voltage:
\[ X_L = 20\sqrt{3} \, \Omega \]
\[ Z = \sqrt{X_L^2 + R^2} = \sqrt{(20\sqrt{3})^2 + 20^2} = \sqrt{1200 + 400} = 40 \, \Omega \]
Power dissipated in the circuit:
\[ P = I_\text{rms}^2 R = \left( \frac{V_\text{rms}}{Z} \right)^2 \times R \]
\[ P = \left( \frac{200}{\sqrt{2} \times 40} \right)^2 \times 20 \]
\[ P = \left( \frac{200}{40\sqrt{2}} \right)^2 \times 20 = 250 \, \text{W} \]
AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is:
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to: