Question

The circuit shown in the figure (Figure 1) uses a neon-filled tube. This neon lamp has a threshold voltage V0 for conduction, because no current flows until the neon gas in the tube is ionized by a sufficiently strong electric field. Once the threshold voltage is exceeded, the lamp has negligible resistance. The capacitor stores electrical energy, which can be released to flash the lamp. Assume that C=0.250μF, R=3.00×106Ω, V0=87.0Vand E=105V.-LC Neon lampA) Assuming that the circuit is hooked up to the emf at time t=0, at what time will the light first flash?B) If the value of R is increased, will the time you found in part A increase or decrease? (select one)The time will not change.The time will increase.The time will decrease
C) The flashing of the lamp is very brief. Why? (select one)
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.
The capacitor discharges through a very high resistance (the lamp filled with ionized gas), and so the discharge time constant is very large. Thus the flash is very brief.
The capacitor discharges through a very low resistance (the lamp filled with ionized gas), and so the discharge time constant is very large. Thus the flash is very brief.
The capacitor discharges through a very high resistance (the lamp filled with ionized gas), and so the discharge time constant is very short. Thus the flash is very brief.
D) Explain what happens after the lamp flashes for the first time. (select one)
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.
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" and remain this state until somebody connect a new battery.
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 decreases, and the capacitor starts to recharge. It charges again and the process will repeat.
None of the above.

Answer 1

Part 1)

Answer:

Explanation:

As we know by equation of charging of the capacitor we will have

$V = E(1 - e^{-t/RC})$

so we will have

$87 = 105(1 - e^{-t/RC})$

here we know that

$R = 3.00 \times 10^6 ohm$

$C = 0.250 \mu F$

so we have

$t = 1.32 s$

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