Physics, 08.10.2020 07:01 kaleahearly123

You take a couple of capacitors and connect them in series, to which you observe a total capacitance of 4.8 µF. However, when you connect them in parallel, their combined capacitance is 35 µF.
Determine the value of each capacitor

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Answer from: livdance

Approximately $\rm 5.7\; \mu F$ and approximately $29\; \rm \mu F$ .

Explanation:

Let C_1 and C_2 denote the capacitance of these two capacitors.

When these two capacitors are connected in parallel, the combined capacitance will be the sum of C_1 and C_2 . (Think about how connecting these two capacitors in parallel is like adding to the total area of the capacitor plates. That would allow a greater amount of charge to be stored.)

$C(\text{parallel}) = C_1 + C_2$ .

On the other hand, when these two capacitors are connected in series, the combined capacitance should satisfy:

$\displaystyle \frac{1}{C(\text{series})} = \frac{1}{C_1} + \frac{1}{C_2}$ .

(Consider how connecting these two capacitors in series is similar to increasing the distance between the capacitor plates. The strength of the electric field () between these plates will become smaller. That translates to a smaller capacitance if the amount of charge stored (Q) stays the same.)

The question states that:

$C(\text{parallel}) = 35\; \rm \mu F$ , and $C(\text{series}) = 4.8\; \rm \mu F$ .

Let the capacitance of these two capacitors be $x\; \rm \mu F$ and $y\; \rm \mu F$ . The two equations will become:

$\displaystyle \left\lbrace \begin{aligned}& x + y = 35 \\ & \frac{1}{x} + \frac{1}{y} = \frac{1}{4.8}\end{aligned}\right.$ .

From the first equation:

y = 35 - x .

Hence, the in the second equation here can be replaced with (35 - x) . That equation would then become:

$\displaystyle \frac{1}{x} + \frac{1}{35 - x} = \frac{1}{4.8}$ .

Solve for :

$\displaystyle \frac{x + (35 - x)}{x \, (35 - x)} = \frac{1}{4.8}$ .

$x\, (35 - x) = 4.8$ .

$x^2 - 35 \, x + 168 = 0$ .

Solve this quadratic equation for :

$x \approx 5.7$ or $x \approx 29.3$ .

Substitute back into the equation y = 35 - x for :

$x \approx 5.7$ and $y \approx 29.3$ , or $x \approx 29.3$ and $x \approx 5.7$ .

In other words, these two capacitors have only one possible set of capacitances (even though the previous quadratic equation gave two distinct real roots.) The capacitances of the two capacitors would be approximately $5.7\; \rm \mu F$ and approximately $29\; \rm \mu F$ (both values are rounded to two significant digits.)

Answer from: Quest

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2 y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x | y

-2 | 2

0 | 8

2 | 14

Create a table of x and y values for the expression y/4=3/4x+2. use the x values of -2,0, and 2. i n

Answer from: Quest

the gas collects in coils that then becomes a liquid that then flows through the coils inside the refrigerator

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You take a couple of capacitors and connect them in series, to which you observe a total capacitance...