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:
Since both branches are identical, the phase difference between \( V_A \) and \( V_B \) and \( V_{in} \) are the same, but in opposite directions. Thus, the phase difference between \( V_{in} \) and \( V_A \) must be \( 45^\circ \), as \( V_{in} \) and \( |V_A - V_D| \) have a difference of \( 90^\circ \). Now, clearly: \[ |R| = (xc) \] Given: \[ 100 \times 10^3 = \frac{10^{12}}{w \times 100} \] Solving for \(w\): \[ w = 10^5 \, \text{rad/s} \] \[ \boxed{w = 10^5 \, \text{rad/s}} \]
An alternating current is represented by the equation, $\mathrm{i}=100 \sqrt{2} \sin (100 \pi \mathrm{t})$ ampere. The RMS value of current and the frequency of the given alternating current are
The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature \( R = 2 \, \text{m} \). Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the side view mirror is \( a \). The value of \( 100a \) is _____________ m/s\(^2\).
Let \( a_1, a_2, a_3, \ldots \) be in an A.P. such that \[ \sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1, \quad a_1 \neq 0. \] If \[ \sum_{k=1}^{n} a_k = 0, \] then \( n \) is: