Question

what is the effective capacitence for this circuit (diagram in attachment)​

Answer 1

Answer

Equivalent capacitance = 48μF

Given

Capacitors 8μF , 8μF , 12μF , 12μF

To Find

Effective capacitance

Solution

If capacitors are connected in parallel ,

Ceq = C1 + C2 + C3 + ...

If capacitors are connected in series ,

1/Ceq = 1/C1 + 1/C2 + 1/C3 + ...

Here , 8μF and 8μF are connected in parallel ,

⇒ Cp = 8 + 8

⇒ Cp = 16μF

Now 8μF , 8μF are connected in series with 12μF and 12μF ,

⇒ 1/Ceq = 1/Cp + 1/12+ 1/12

⇒ 1/Ceq = 1/16 + 1/12 + 1/12

⇒ 1/Ceq = 1/16 + 1/6

⇒ 1/Ceq = 1/48

⇒ Ceq = 48μF

So , Equivalent capacitance = 48μF

Answer 2

Answer:

Answer

Equivalent capacitance = 48μF

Given

Capacitors 8μF , 8μF , 12μF , 12μF

To Find

Effective capacitance

Solution

If capacitors are connected in parallel ,

Ceq = C1 + C2 + C3 + ...

If capacitors are connected in series ,

1/Ceq = 1/C1 + 1/C2 + 1/C3 + ...

Here , 8μF and 8μF are connected in parallel ,

⇒ Cp = 8 + 8

⇒ Cp = 16μF

Now 8μF , 8μF are connected in series with 12μF and 12μF ,

⇒ 1/Ceq = 1/Cp + 1/12+ 1/12

⇒ 1/Ceq = 1/16 + 1/12 + 1/12

⇒ 1/Ceq = 1/16 + 1/6

⇒ 1/Ceq = 1/48

⇒ Ceq = 48μF

So , Equivalent capacitance = 48μF