Answer :

In the given fig. Capacitors B and C are in parallel.

The formula for parallel combination of capacitor is

C_{eq}=C_{1}+C_{2}= C_{A} +C_{B}= 4 + 4 =8 μF

The formula for series combination of capacitors is

From the figure, the 8 μF is connected in series with C_{eqv}.

Thus, we get

Putting the values in the above equation, we get

Capacitors B and C are in parallel. C_{C}

System of B,C and A has the same capacitor values. So they exhibit the same potential difference between them.

Hence Voltage across A is =6V

The voltage across B and C is = 6V

The charge can be calculated as

Where

Q is the charge of the capacitor

C is the Capacitance of the capacitor

V is the voltage across the circuit

Charge flows through A is Q* _{A}* = 8×6= 48μC

Charge flows through B is Q* _{B}*= 4×6 =24μC

Charge flows through C is Q* _{C}*= 4×6 = 24μC

Charge appearing on the capacitors A, B and C is 48μC, 24μC and 24μC respectively.

Rate this question :

In the circuit shPhysics - Exemplar

Convince yourselfHC Verma - Concepts of Physics Part 2

Each capacitor inHC Verma - Concepts of Physics Part 2

Each capacitor shHC Verma - Concepts of Physics Part 2

A cylindrical capHC Verma - Concepts of Physics Part 2

In the circuit shPhysics - Exemplar

A capacitor is maHC Verma - Concepts of Physics Part 2

Find the equivaleHC Verma - Concepts of Physics Part 2

A capacitor of caHC Verma - Concepts of Physics Part 2

A finite ladder iHC Verma - Concepts of Physics Part 2