HOW CARELESS, A PREVIOUS CORRECT RESPONSE FROM ROTTEN STUDENT1536 WAS INADVERTEN

Question

HOW CARELESS, A PREVIOUS CORRECT RESPONSE FROM ROTTEN STUDENT1536 WAS INADVERTENTLY DELETED. PLEASE SHOW ALL STEPS: A 40F capacitor is in series with a 5000 OHM resistor; the resistor is charged by a 100V battery. A neon lamp is connected in parallel with the capacitor. When the voltage across the lamp reaches 70V it fires and its resistance goes to zero almost instantaneously. It then ceases to conduct once the votage drops below 40V, so the capacitor again begins charging back up to 70V then it discharges again...This is the kind of circuit used to make roadside warning lights at construction sites. Determine the frequency at which the light will blink

Explanation / Answer

vr=5000xi

vc=100-5000i

40=dq/dv

40=i/(dv/dt)

i=(100-v)/5000

substitute in above equation

dv/dt=(100-v)/(20x10^4)

dv/(100-v)=dt/(20x10^4)

solving this we get

log(100-v)=t/(20x10^4)+c

now substitute the boundary values as v1=40v to v2=70v

and time from t1=0 to t2=t

then t= log(2)x20x10^4

t=6.0205x10^4 secs

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