Answer:
Resistance in the flash tube, [tex]R=3.97\times 10^{-3}\ \Omega[/tex]
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, [tex]C=503\ \mu F=503\times 10^{-6}\ F[/tex]
The time constant of RC circuit is given by :
[tex]\tau=RC[/tex]
R is the resistance in the flash tube
[tex]R=\dfrac{\tau}{C}[/tex]..........(1)
Speed of the bullet is given by total distance divided by total time taken as :
[tex]v=\dfrac{d}{\tau}[/tex]
[tex]\tau=\dfrac{d}{v}[/tex]
[tex]\tau=\dfrac{0.001}{500}[/tex]
[tex]\tau=0.000002\ s[/tex]
Equation (1) becomes :
[tex]R=\dfrac{0.000002}{503\times 10^{-6}}[/tex]
[tex]R=3.97\times 10^{-3}\ \Omega[/tex]
So, the resistance in the flash tube is [tex]3.97\times 10^{-3}\ \Omega[/tex]. Hence, this is the required solution.
Answer:
[tex]R=3.98\times 10^{-3} ohm[/tex]
Explanation:
We are given that
Speed of bullet=500 m/s
Distance during one RC time constant=d=1 mm=[tex]1\times 10^{-3} m[/tex]
Capacitance=[tex]503\mu F=503\times 10^{-6} F[/tex]
We have to find the resistance in the flash tube.
[tex]\tau=\frac{d}{v}=\frac{1\times 10^{-3}}{500}[/tex]
[tex]\tau=2\times 10^{-6} s[/tex]
[tex]\tau =RC[/tex]
[tex]R=\frac{\tau}{C}=\frac{2\times 10^{-6}}{503\times 10^{-6}}[/tex]
[tex]R=3.98\times 10^{-3} ohm[/tex]
Hence, the resistance in the flash tube=[tex]3.98\times 10^{-3} ohm[/tex]
