Part 1)
Answer:
Explanation:
As we know by equation of charging of the capacitor we will have
so we will have
here we know that
so we have
Part b)
Answer:
The time will increase.
Explanation:
As we know that on increasing the value of the resistance the the product of the resistance and capacitance will increase so the time will increase to get the above voltage.
Part c)
Answer:
The capacitor discharges through a very low resistance (the lamp filled with ionized gas), and so the discharge time constant is very short. Thus the flash is very brief.
Explanation:
Since the lamp resistance is very small so the energy across the lamp will totally lost in very short interval of time
Part d)
Answer:
Once the lamp has flashed, the stored energy in the capacitor is gone, and there is no source of charge to maintain the lamp current. The lamp "goes out", the lamp resistance increases, and the capacitor starts to recharge. It charges again and the process will repeat.
Explanation:
Since we know that the battery is connected to the given system so after whole energy of capacitor is flashed out it is again charged by the battery and the process will continue