Question 19.SP.4: A RL circuit (Fig. 19-9) is used to generate a high voltage ......
A RL circuit (Fig. 19-9) is used to generate a high voltage to light a neon bulb, which requires 90 V for ionization at which time it glows. When the circuit is open, the high resistance R_{2} produces a low time constant L/R so that the current drops toward zero much faster than the rise of current when the switch is closed. The result is a high value of self-induced voltage across a coil when the RL circuit is open. This voltage can be greater than the applied voltage. (a) Find the time constants of the circuit when the switch is opened and when the switch is closed. (b) When the switch is closed, find the voltage across the neon bulb. Is it sufficient to start ionization? (c) When the switch is open, what is the voltage across the neon bulb, and is it great enough to light it?
(a) The time constant is the ratio of total inductance to the total resistance.
Switch open:
T=\frac{L}{R_1+R_2}=\frac{2}{4000+100}=\frac{2}{4100}=4.9 \times 10^{-4}=0.49\ msSwitch closed (R_{2} short-circuited):
T=\frac{L}{R_1}=\frac{2}{100}=2 \times 10^{-2}=20\ msWhen the switch is open, there is a much shorter time constant for current decay. The current decays practically to zero after five time constants, or 2.5 ms.
(b) When the switch is closed, there is 10 V across the neon bulb, which is far less than the 90 V needed to ionize it to glow.
(c) When the switch is closed, R_{2} is short-circuited so that the 100-Ω R_{1} is the only resistance. The steady-state current I=V/R_{1} = 10/100 = 0.10 A. When the switch is opened, the rapid drop in current results in a magnetic field collapsing at a fast rate, inducing a high voltage across L. The energy stored in the magnetic field maintains I at 0.10 A for an instant before the current decays. With 0.10 A in the 4-kΩ R_{2}, its potential difference is 0.10(4000) = 400 V. This 400-V pulse is sufficient to light the bulb.