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SVETLANKA909090 [29]

3 years ago

11

N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2

and third one is C/4, forth one is C/8 and so on. Namely, capacitance of a capacitor is one-half of the previous one. What is the equivalent capacitance of this parallel combination when N goes to inifinity?

1 answer:

liberstina [14]3 years ago

5 0

Answer:

2C

Explanation:

The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.

So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.

Using the formula for the sum of the infinite terms of a geometric series, we have:

Sum = First term / (1 - rate)

Sum = C / (1 - 0.5)

Sum = C / 0.5 = 2C

So the equivalent capacitance of this parallel connection is 2C.

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Answer:

a

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b

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Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

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$Tsin \theta = \frac{mv^2}{r}$

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$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

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The horizontal component is mathematically represented as

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substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

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substituting value

$T_ y = 0.344 * 9.8$

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The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

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bezimeni [28]

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