Question

If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge through a flash tube can limit blurring. Assuming 1.00 mm of motion during one RC constant is acceptable, and given that the flash is driven by a 503 µF capacitor, what is the resistance in the flash tube?

Answer 1

`R=3.98* 10^(-3) ohm`

Step-by-step explanation:

We are given that

Speed of bullet=500 m/s

Distance during one RC time constant=d=1 mm=
`1* 10^(-3) m`

Capacitance=
`503\mu F=503* 10^(-6) F`

We have to find the resistance in the flash tube.

`\tau=(d)/(v)=(1* 10^(-3))/(500)`

`\tau=2* 10^(-6) s`

`\tau =RC`

`R=(\tau)/(C)=(2* 10^(-6))/(503* 10^(-6))`

`R=3.98* 10^(-3) ohm`

Hence, the resistance in the flash tube=
`3.98* 10^(-3) ohm`

Answer 2

Resistance in the flash tube,
`R=3.97* 10^(-3)\ \Omega`

Step-by-step explanation:

It is given that,

Speed of the bullet, v = 500 m/s

Distance between one RC constant, d = 1 mm = 0.001 m

Capacitance,
`C=503\ \mu F=503* 10^(-6)\ F`

The time constant of RC circuit is given by :

`\tau=RC`

R is the resistance in the flash tube

`R=(\tau)/(C)`..........(1)

Speed of the bullet is given by total distance divided by total time taken as :

`v=(d)/(\tau)`

`\tau=(d)/(v)`

`\tau=(0.001)/(500)`

`\tau=0.000002\ s`

Equation (1) becomes :

`R=(0.000002)/(503* 10^(-6))`

`R=3.97* 10^(-3)\ \Omega`

So, the resistance in the flash tube is
`3.97* 10^(-3)\ \Omega`. Hence, this is the required solution.