Question

in the circuit the potential difference between a and b is ?​

Answer 1

Given :

In the circuit voltage source = 20 V

Values of capacitance of each capacitor is shown in the diagram.

To find :

Potential Difference between  A and B

Solution :

Draw the equivalent circuit diagram by calculating the equivalent capacitance of the circuit.

capacitors in series =

 $\frac{1}{C} =\frac{1}{C_{1} } +\frac{1}{C_{2} }$

capacitors in parallel=

$C = C_{1} + C_{2}$

2 capacitors magnitude 8uF are in series . equivalent capacitance = 4uF

2 capacitors with capacitance 4uF are parallel, equivalent capacitance = 8uF

apply kirchoff's law and V =$\frac{Q}{C}$

Q = CV

equivalent capacitance = 1.6 uF

Q = 1.6(20)

Q = 32C

voltage drop across 2uF capacitor = Q/C

                                                          32/2 = 16 volts.

boltage drop across capacitor of capacitance 4uF is 4V

Charge will be equally divided in both the branches since the value of capacitance is same = 4uF.

charge in each branch = 16C.

voltage drop across 8uF capacitor = Q/C

                                                          16/8 = 2 volts

hence $V_{AB = 2 volts$

Potential difference between A and B is 2 volts.