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laila [671]
2 years ago
6

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T

he capacitor is charged to 5.03 V, then discharged through the resistor. The time for the potential difference across the plates to decrease to 0.833 V is measured by a clock inside the game. If the range of discharge times that can be handled effectively is from 11.5 μs to 6.33 ms, what should be the (a) lower value and (b) higher value of the resistance range of the resistor?
Engineering
1 answer:
anyanavicka [17]2 years ago
4 0

Answer:

R min = 28.173 ohm

R max = 1.55 × 10^{4}  ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates =  0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = V_o \times e^{t\RC}    ...........1

so R will be

R = \frac{t}{C\times ln(\frac{V_o}{V})}    ....................2

put here value

so for t min 11.5 μs

R = \frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}

R min = 28.173 ohm

and

for t max 6.33 ms

R max = \frac{6.33}{11.5} \times 28.173  

R max = 1.55 × 10^{4}  ohm

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