In Figure 21.41, does the graph indicate the time constant is shorter for discharging than for charging? Would you expect ionized gas to have low resistance? How would you adjust R to get a longer time between flashes? Would adjusting R affect the discharge time?

The discharging of the capacitor is,

\({\rm{V = }}{{\rm{V}}_{\rm{0}}}{{\rm{e}}^{\frac{{{\rm{ - t}}}}{{{\rm{RC}}}}}}\)

Here,\({{\rm{V}}_{\rm{0}}}\)is the voltage across the capacitor when it is charged to the maximum,\({\rm{t}}\)is the time from the initiation of discharging,\({\rm{R}}\)is the resistance in the circuit, and\({\rm{C}}\)is the value of capacitance of the capacitor.

In this situation, the discharge of a capacitor is faster than the charging process. This is because when the light bulb is shining and heating up, it has a high resistance, which increases as the temperature rises. The voltage across the light bulb reduces as the capacitor discharges, causing it to dim and cool. The resistance, and hence the time constant RC, diminishes.

Due to the high kinetic energy of the ions, it is more difficult to guide the flow of current in an ionized gas at a higher temperature. To acquire a longer period between flashes, the charging time must be increased so that the light remains on for longer. R is increased to do this. This does not affect the discharge time, which is solely determined by the light bulb's resistance.