Resistor and Capacitor Arrangements

Question

Three capacitors are connected as shown in the figure. What is the equivalent capacitance between points A and B? Please explain the steps leading to your answer.

If you were to swap all capacitors in this circuit with resistors of similar respective magnitudes in units of Ω, what would the equivalent resistor be? Please show details of your reasoning.

Answer 1

This is the equivalent resistance answer.

For two resistors in series, the equivalent resistance is just the sum or the two resistances:

An R₁ Ω resistor in series with an R₂ Ω resistor leads to an equivalent resistance R_s = (R₁ + R₂) Ω.

For two resistors in parallel, the reciprocal of the equivalent resistance is the sum or the reciprocals of the two resistances:

An R₁ Ω resistor in parallel with an R₂ Ω resistor leads to an equivalent resistance given by

1/R_p = 1/R₁ +1/R₂

This leads, after some algebraic manipulation, to the formula

R_p = (R₁)(R₂)/(R₁ + R₂) Ω

The 4 Ω and 8 Ω resistors in series are equivalent to a 12 Ω resistor.

This makes middle part (between the junctions) equivalent to 6 Ω and 12 Ω parallel resistors.

So the equivalent resistance of the three resistors between the junctions is given by

(6)(12)/(6 + 12) = 72/18 = 4 Ω

Finally, we have a 12 Ω on top in series with a group of resistors equivalent to a 4 Ω resistor, which makes the entire group equivalent to a 16 Ω resistor.

Answer

16 Ω

When analyzing circuits with multiple capacitors, the series and parallel rules are reversed:

For two capacitors in parallel, the equivalent capacitance is just the sum or the two capacitances.

For two capacitors in series, the reciprocal of the equivalent capacitance is the sum or the reciprocals of the two capacitances.