The equivalent capacitance of the circuit given between A and B is 
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:
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
A point particle of charge \( Q \) is located at \( P \) along the axis of an electric dipole 1 at a distance \( r \) as shown in the figure. The point \( P \) is also on the equatorial plane of a second electric dipole 2 at a distance \( r \). The dipoles are made of opposite charge \( q \) separated by a distance \( 2a \). For the charge particle at \( P \) not to experience any net force, which of the following correctly describes the situation?

The total capacitance of this equivalent single capacitor depends both on the individual capacitors and how they are connected. There are two simple and common types of connections, called series and parallel, for which we can easily calculate the total capacitance.
Read Also: Combination of Capacitors
When one terminal of a capacitor is connected to the terminal of another capacitors , called series combination of capacitors.
Capacitors can be connected in two types which are in series and in parallel. If capacitors are connected one after the other in the form of a chain then it is in series. In series, the capacitance is less.
When the capacitors are connected between two common points they are called to be connected in parallel.
When the plates are connected in parallel the size of the plates gets doubled, because of that the capacitance is doubled. So in a parallel combination of capacitors, we get more capacitance.
Read More: Types of Capacitors