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A circuit contains a D cell battery, a switch, an 83-Ω resistor, and three 83-F capacitors connected in series.

(a) What is the equivalent capacitance of the circuit (in F)?

Answer :

Final answer:

The equivalent capacitance of the series circuit containing three 83-F capacitors is 27.67 F. The equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances.

Explanation:

To find the equivalent capacitance of the circuit in series.

In this case, you have 3 capacitors each of 83 F.

So, the equivalent capacitance (Ceq) would be calculated as shown:

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

Substituting the values, we get 1/Ceq = 1/83 + 1/83 + 1/83 = 3/83.

Therefore, Ceq = 83/3, which simplifies to 27.67 F.

To find the equivalent capacitance of capacitors connected in series, you can use the formula:

1 / C_eq = 1 / C_1 + 1 / C_2 + 1 / C_3 + ...

In your case, you have three 83-µF capacitors connected in series.

So, you can calculate the equivalent capacitance as follows:

1 / C_eq = 1 / (83 µF) + 1 / (83 µF) + 1 / (83 µF)

Now, add these inverses:

1 / C_eq = (1 / 83 + 1 / 83 + 1 / 83) µF^(-1)

1 / C_eq = (3 / 83) µF^(-1)

Now, take the reciprocal of both sides to find C_eq:

C_eq = 83 / 3 µF ≈ 27.67 µF

So, the equivalent capacitance of the circuit is approximately 27.67 µF.

Learn more about Equivalent Capacitance

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