Question

foe the circuit shown here, the potential difference between points A and B
. plzz give me full solution of this and I swear I will follow you ..​

Answer 1

answer is 4.7 volt

in first picture i simplify the circuit so, that it should be easy . in second picture i find the value of Q and then put into the equation of Va-Vb .

there is no use of resistance .

Answer 2

Answer:

For the circuit shown, the potential difference between A and B is 2.5V.

Explanation:

For the circuit shown, to find the potential difference between A and B,

1. Solve and reduce the circuit as much as possible.
2. Remove any redundant elements from the circuit.
3. Apply KCL/KVL current or voltage division formulae on the circuit according to the value to be determined.

Calculation:

• Reduce the two series capacitors on the right 2C, 2C into a single capacitor.

       C' = (2C * 2C)/(2C + 2C)

       C' = C  

• Reduce the two parallel capacitors on the right C, C into a single capacitor.

      C' = C + C = 2C

• Reduce the two series capacitors on the left 6C, 6C into a single capacitor.

      C' = (6C * 6C)/(6C + 6C)

       C' = 3C  

• Reduce the two parallel capacitors on the left 3C, 3C into a single capacitor.

        C' = 3C + 3C = 6C

• Remove the resistor and the capacitor C parallel to the voltage source as they are redundant.
• Now we have a flow in the circuit as A, 6C, B, 2C, V in clockwise direction.

   V = Q/C

   Vab = V - V_2C

   V_2C = Q/2C

   C_eq = (2C * 6C)/(2C + 6C) = 1.5C

    Q = 10 * C_eq = 10 * 1.5C = 15C

     Vab = 10 - 15C/2C = 10 - 7.5

The potential difference between A and B is 2.5V.

Know more about capacitors connections:

https://brainly.in/question/11048536

https://brainly.in/question/21653924