In the following circuit C1 41 F C2 2 41 F C3 2 41 F C4

In the following circuit. C_1 = 4.1 F, C_2 = 2* 4.1 F, C_3 = 2 * 4.1 F, C_4 = 2* 6.0 F, and c_5 = 3* 6.0. Calculate the equivalent capacitance in the circuit. [Picture F Five Caps.jpg]

Solution

The problem requires one to identify the pairs of capacitors that are in series and in parallel and sequentially reduce each pair to a single equivalent capacitor.

Lets begin.

Note that C2 and C3 are parallel to each other. So then they can be reduced to an equivalent capacitor of capacitance C\' = C2 + C3 = 4*4.1 F

Now this C\' and C1 are in series. This can be reduced to a single capacitance C\'\' = C\'*C1/(C\' + C1) = 0.8*4.1 F

C4 and C5 are in series. So, they can be reduced to a single capacitor C\'\'\' = C4*C5/(C4 + C5) = 1.2 * 6.0 F

Now the circuit is reduced to C\'\' and C\'\'\' in parallel.

The equivalent capacitance of that system is given by C_eq = C\'\' + C\'\'\' = .8*4.1 + 1.2*6.0 = 10.48 F

So the answer is 10.48 Farads