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Stells [14]
3 years ago
15

The capacitor in the flash of a disposable camera has a value of 165 μF. 1) What is the resistance of the filament in the bulb i

f it takes 10 s to charge the capacitor to 80% of its maximum charge? (Express your answer to two significant figures.)
Physics
1 answer:
pashok25 [27]3 years ago
7 0

Answer:

3.8 x 10⁴ Ω

Explanation:

C = Capacitance = 165 μF

R = resistance of the filament = ?

t = time taken to charge the capacitor = 10 s

Q₀ = maximum charge stored by capacitor

Q = charge stored by capacitor at time "t" = 0.80 Q₀

T = Time constant

Charge stored by capacitor at any time is given as

Q = Q_{o}(1 - e^{\frac{-t}{T}})

0.80 Q_{o} = Q_{o}(1 - e^{\frac{-10}{T}})

T = 6.21 s

Time constant is given as

T = RC

6.21 = R (165 x 10⁻⁶)

R = 3.8 x 10⁴ Ω

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of n is n =7

Explanation:

    From the question we are told that

          The value of m = 2

            For every value of m, n = m+ 1, m+2,m+3,....

           The modified version of  Balmer's formula is \frac{1}{\lambda}  = R [\frac{1}{m^2} - \frac{1}{n^2}  ]

             The Rydberg constant has a value of R = 1.097 *10^{7} m^{-1}

The objective of this solution is to obtain the value of n for which the wavelength of the Balmer series line is smaller than 400nm

   

For m = 2 and n =3

    The wavelength is

                          \frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{3^2}  ]

                          \lambda = \frac{1}{1523611.1112}

                             \lambda = 656nm

For m = 2 and n = 4

    The wavelength is

                          \frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{4^2}  ]

                          \lambda = \frac{1}{2056875}

                             \lambda = 486nm

For m = 2 and n = 5

    The wavelength is

                          \frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{5^2}  ]

                          \lambda = \frac{1}{2303700}

                             \lambda = 434nm

For m = 2 and n = 6

    The wavelength is

                          \frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{6^2}  ]

                          \lambda = \frac{1}{2422222}

                             \lambda = 410nm

For m = 2 and n = 7

    The wavelength is

                          \frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{7^2}  ]

                          \lambda = \frac{1}{2518622}

                             \lambda = 397nm

So the value of n is  7

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