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iVinArrow [24]
2 years ago
8

What determines how long it takes for the capacitor to charge?

Physics
1 answer:
myrzilka [38]2 years ago
6 0

The time constant determines how long it takes for the capacitor to charge.

To find the answer, we have to know more about the time constant of the capacitor.

<h3>What is time constant?</h3>
  • The time it takes for a capacitor to discharge 36.8% of its charge in a discharging circuit or charge up to 63.2% of its maximum capacity in a charging circuit, given that it has no initial charge, is the time constant of a resistor-capacitor series combination.
  • The circuit's reaction to a step-up (or constant) voltage input is likewise determined by the time constant.
  • As a result, the time constant determines the circuit's cutoff frequency.

Thus, we can conclude that, the time constant determines how long it takes for the capacitor to charge.

Learn more about the time constant here:

brainly.com/question/17050299

#SPJ4

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Explanation:

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you drop a 2 kg book to a friend who stand on the ground at distance D=10.0 m below, if your friends out streched hand at distan
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n noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the positio
lys-0071 [83]

Answer:

7.888m

Explanation:

Given:

frequency 'f'= 90Hz

velocity of the sound 'v' = 340 m /sec

The wavelength of the wave is given by,

λ= v/f => 340/90

λ= 3.777m

The destructive interference condition is givn by

Δd= (m+\frac{1}{2}  ) λ

where, m=0,1,2,3,..

m=0, for minimum destructive interference

Δd= (0+\frac{1}{2}  ) x 3.777

Δd=1.888

Therefore, the required distance is

d_f=d_i + Δd => 6 + 1.888

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Thus, So the speaker should be placed at  7.888 m

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A racemic mixture can rotate a plane-polarized light in a clockwise direction. Select one: True False
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Answer:

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Explanation:

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3 years ago
A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it
mojhsa [17]

Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm

r = 0.145\ cm = 0.145\times 10^{-2}\ m

Charge ,

Q = 30.0\ pC=30\times 10^{-12}

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}

Where,  

V= Electric potential,  

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge  

r = Radius  

Substituting the values we get

V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}

V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt

Therefore,

The potential (in V) near its surface is 186.13 Volt.

3 0
4 years ago
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