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Find the total capacitance of the circuit if all capacitors are 4uF.

Find The Total Capacitance Of The Circuit If All Capacitors Are 4uF class=

Sagot :

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Start with the left part (3 capacitors). The two capacitors that forms an L-shape are in series, so their capacitance is 1/(1/4 + 1/4) = 2μF. The remaining capacitors in the left part are in parallel, so their capacitance is

2 μF + 4 μF = 6 μF.

Next, calculate the capacitance of the right part (3 capacitors). This is similar to the left part, and the capacitance of the right part is 6 μF as well.

Calculate the capacitance of the top part (2 capacitors). They are in parallel, so the capacitance of the top part is 4μF + 4μF = 8μF.

The left, right, and top part are in series. So the total capacitance of the circuit is 1/(1/6 + 1/6 + 1/8) = 24/11 μF or 2.18 μF.