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

Answer:

Resistance in the flash tube, $R=3.97\times 10^{-3}\ \Omega$

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, $C=503\ \mu F=503\times 10^{-6}\ F$

The time constant of RC circuit is given by :

$\tau=RC$

R is the resistance in the flash tube

$R=\dfrac{\tau}{C}$..........(1)

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

$v=\dfrac{d}{\tau}$

$\tau=\dfrac{d}{v}$

$\tau=\dfrac{0.001}{500}$

$\tau=0.000002\ s$

Equation (1) becomes :

$R=\dfrac{0.000002}{503\times 10^{-6}}$

$R=3.97\times 10^{-3}\ \Omega$

So, the resistance in the flash tube is $3.97\times 10^{-3}\ \Omega$. Hence, this is the required solution.

Answer 2

Answer:

$R=3.98\times 10^{-3} ohm$

Explanation:

We are given that

Speed of bullet=500 m/s

Distance during one  RC time constant=d=1 mm=$1\times 10^{-3} m$

Capacitance=$503\mu F=503\times 10^{-6} F$

We have to find the resistance in the flash tube.

$\tau=\frac{d}{v}=\frac{1\times 10^{-3}}{500}$

$\tau=2\times 10^{-6} s$

$\tau =RC$

$R=\frac{\tau}{C}=\frac{2\times 10^{-6}}{503\times 10^{-6}}$

$R=3.98\times 10^{-3} ohm$

Hence, the resistance in the flash tube=$3.98\times 10^{-3} ohm$