Capacitor

Once a capacitor is fully charged, it behaves like an open circuit for DC current. So, we can remove all the capacitors to simplify the circuit and then obtain potential difference across .On simplifying we get the following circuit which can be remade as the following:$ F B H B 2 k \Omega Q=C V $

To analyze this circuit, let's start by understanding the initial scenario. Capacitor with a capacitance of is connected to a battery with a potential . The charge on is calculated as: After is disconnected from the battery, it is connected in parallel with another capacitor, , which has a capacitance of . Since they are connected in parallel, they will share the charge while maintaining the same voltage across them. The total charge initially on is conserved and distributed between both capacitors.The initial total charge: After the connection, let and denote the charges on and , respectively. In a parallel setup, the voltage across both capacitors is the same, denoted as : With charge conservation, the equation is: Substituting the expressions for and : Solving for : Now, to find the ratio , we calculate using : The total charge is maintained as: Thus, the ratio becomes: This simplifies to: Substituting the given capacitance values: Therefore, the ratio is:

Step 1: Understand the networkThe four capacitors form adiamond (bridge) networkbetween the left and right terminals of a12 V battery.Top-left capacitor =4 μFTop-right capacitor =2 μFBottom-left capacitor =2 μFBottom-right capacitor =4 μFPointPis the junction of the top capacitors.PointQis the junction of the bottom capacitors.Step 2: Voltage division in each branchEach branch behaves like aseries combination:Top branch (4 μF and 2 μF in series)✅Answer:4V ✔️