Question

Determine the equivalent capacitance of the combination shown when C = 12 n

Answer 1

Hi there!

Recall the following.

Capacitors in series:

`C_T = (1)/(C_1) + (1)/(C_2) +...+ (1)/(C_n)`

Capacitors in parallel:

`C_T = C_1 + C_2 + ... + C_n`

Begin by solving for the resulting capacitance of both paths.

Path on the left:

`(1)/(C_T) = (1)/(2C) + (1)/(C) = (3)/(2C)\\C_T = (2C)/(3)`

Path on the right:

`(1)/(C_T) = (1)/(C) + (1)/(3C) = (4)/(3C)\\\\C_T = (3C)/(4)`

Now, since we ADD capacitors in parallel, we can add the resulting capacitances together:

`C_T = (2C)/(3) + (3C)/(4)`

Substitute in 12 F for C and solve.

`C_T = (2(12))/(3) + (3(12))/(4) = (24)/(3) + (36)/(4) = 8 + 9 = \boxed{17F}`