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Lesechka [4]
3 years ago
10

Lucas plugs a lamp into the wall. The light bulb in the lamp turns on. What kind of energy transformation is this

Physics
2 answers:
Katena32 [7]3 years ago
7 0
A light bulb converts electric energy into heat and light energy, but it is different when you are using solar energy.Not that different though, the solar energy turns into electric energy that converts into heat and light energy.
pantera1 [17]3 years ago
5 0

Answer:

Electric energy from the outlet transforms to radiant energy and thermal energy in the lightbulb.

Explanation:

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Answer:

Diffraction.

Explanation:

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You take a couple of capacitors and connect them in series, to which you observe a total capacitance
Zepler [3.9K]

Answer:

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 (V) 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 y 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 x:

\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:

x \approx 5.7 or x \approx 29.3.

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

  • 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.)

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The higher the phase margin the more stable is the system and for these tuned parameters, the phase margin is around
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