According to its design specification, the timer circuit delaying the closing of an elevator door is to have a capacitance of 30.0 µF between two points A and B.

Question

According to its design specification, the timer circuit delaying the closing of an elevator door is to have a capacitance of 30.0 µF between two points A and B.

(a) When one circuit is being constructed, the inexpensive but durable capacitor installed between these two points is found to have capacitance 33.6 µF. To meet the specification, one additional capacitor can be placed between the two points. Should it be in series or in parallel with the 33.6 µF capacitor?
in series
in parallel

What should be its capacitance?
µF
(b) The next circuit comes down the assembly line with capacitance 29.3 µF between A and B. What additional capacitor should be installed in series or in parallel in that circuit, to meet the specification?
in series
in parallel

µF

Answer 1

Capacitors in series subract.

Capacitors in parallel add.

#2.
!/c = 1/c1 + 1/c2
You know c and c2, substitute and solve for c2.

Answer 2

To solve this problem, we need to understand the concept of capacitance in series and parallel connections.

In a series connection, the capacitance of the overall circuit is given by the reciprocal sum of the individual capacitances:

1/C_series = 1/C1 + 1/C2

In a parallel connection, the capacitance of the overall circuit is the sum of the individual capacitances:

C_parallel = C1 + C2

(a) In this case, the existing capacitor has a capacitance of 33.6 µF, and we need to add one more capacitor to meet the target capacitance of 30.0 µF. Based on the options given, we need to find the capacitance that, when connected in series with the existing capacitor, will give a total capacitance of 30.0 µF.

Let's assume the additional capacitor has a capacitance of C. So, using the series capacitance formula, we have:

1/30.0 µF = 1/33.6 µF + 1/C

To find C, we can solve this equation. Rearranging the equation, we get:

1/C = 1/30.0 µF - 1/33.6 µF

Calculating the values on the right side of the equation, we have:

1/C = (33.6 - 30.0)/(30.0 * 33.6)

Simplifying:

1/C = 3.6/1008.0

1/C = 0.003571

Now, taking the reciprocal of both sides:

C = 1/0.003571

C ≈ 280 µF

Therefore, the additional capacitor should have a capacitance of approximately 280 µF, and it needs to be connected in series with the existing 33.6 µF capacitor.

(b) In this case, the existing capacitor has a capacitance of 29.3 µF, and we need to add another capacitor to meet the target capacitance of 30.0 µF. Based on the options given, we need to find the capacitance that, when connected in series or in parallel with the existing capacitor, will give a total capacitance of 30.0 µF.

Using the same approach as before, we can set up the equation:

1/30.0 µF = 1/29.3 µF + 1/C

To determine whether the additional capacitor should be connected in series or parallel, we need to compare the values of C. For series connection, we calculate C using the series capacitance formula. For parallel connection, we calculate C using the parallel capacitance formula:

Series case:
1/C = 1/30.0 µF - 1/29.3 µF

Parallel case:
C = 30.0 µF - 29.3 µF

Solving these equations, we find:

Series case: C ≈ 1630 µF

Parallel case: C ≈ 0.7 µF

Comparing the calculated values, we can see that the capacitance of 0.7 µF is the closest to the required additional capacitance. Therefore, the additional capacitor should have a capacitance of approximately 0.7 µF, and it needs to be connected in parallel with the existing 29.3 µF capacitor.