What should be the capacity of capacitor of r-c circuit, in the value of resistance is 10 to become the value of time constant 10?
(a) 10uf
(b) 100uf
(c) 1000uf
(d) 10,000uf

☆During discharge the current though resistance is i=i0e−t/η where η= time constant.

Power dissipation through resistor R is P=i2R=i02e−2t/ηR=P0e−t/(η/2) 

Thus time constant becomes 2η

✌✌

answer:

Physically, the time constant represents the elapsed time required for the system response to decay to zero if the system had continued to decay at the initial rate, because of the progressive change in the rate of decay the response will have actually decreased in value to 1 / e ≈ 36.8% in this time (say from a step ...

Explanation:

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C m 2.158 is yor correct answer

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Explanation:

Explanation:

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Explanation:

Physically, the time constant represents the elapsed time required for the system response to decay to zero if the system had continued to decay at the initial rate, because of the progressive change in the rate of decay the response will have actually decreased in value to 1 / e ≈ 36.8% in this time (say from a step ...

Given :-

Time constant = η

To Find :-

The value of time constant for the power dissipated

Solution:-

We know that the current

Power P = i²R

Now putting the value of i in above equation.

P =

P =

P =     Where  

Then the value of time constant for power dissipated through the resistance will be η/2 i.e  time constant will be

hewudgh है i rt t ye it uti uuytu he

Explanation:

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known :

qa = 3x10⁻⁸ c

qb = -6x10⁻⁸ c

qc = 2x10⁻⁸ c

rab = 3 cm = 3x10⁻² m

rbc = 6 cm = 6x10⁻² m

asked: f at point b?

answered:

coulomb force of charge a and b

fab = k qaqb / rab²

= 9x10⁹ x 3x10⁻⁸ x 6x10⁻⁸ / (3x10⁻²) ²

= 9x10⁹ x 18x10⁻¹⁶ / 9x10⁻⁴

= 18x10⁻³ n

coulomb force charge c and b

fbc = k qbqc / rbc²

= 9x10⁹ x 6x10⁻⁸ x 2x10⁻⁸ / (6x10⁻²) ²

= 108x10⁻⁷ / 36x10⁻⁴

= 3x10⁻³ n

so the coulomb force at point b = fab - fbc

= 18x10⁻³ - 3x10⁻³

= 15x10⁻³ n

= 1.5x10⁻² n

answer:

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