Item 5 Figure shows the circuit for a simple sawtooth oscillator. At time t = 0,

Question

Item 5 Figure shows the circuit for a simple sawtooth oscillator. At time t = 0, its switch S is closed. The neon bulb has initially infinite resistance until the voltage across it reaches 90.0V , and then it begins to conduct with very little resistance (essentially zero). It stops conducting (its resistance becomes essentially infinite) when the voltage drops down to 70.0V. Part A At what time t1 does the neon bulb reach 90.0V and start conducting? Express your answer with the appropriate units. Part B At what time t2 does the bulb reach 70.0V for a second time and again become conducting? Express your answer with the appropriate units.

Explanation / Answer

Here ,

when charging ,

time constant , T1 = RC

T1 = 33 *10^3 * 4 *10^-6

T1 = 0.132 s

Now , for charging of capacitor ,

V = Vo(1 - e^(-t1/T1))

90 = 100 * (1 - e^(-t1/.132))

solving for t1

t1 = 0.304 s

the time t1 is 0.304 s

b)

as the resistance becomes very small whem conducting , the time tp discharge is very small ,

now , time taken to reach 70 V

70 = 100 * (1 - e^-t/.132)

solving for t

t = 0.16 s

Now ,time taken to reach 90 from 70 = 0.304 - 0.16

time taken to reach 90 from 70 = 0.145 s

therefore , time t2 = t1 + 0.145

t2 = 0.304 + 0.145

t2 = 0.449 s

the time t2 is 0.449 s

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