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2 years ago

10

he timing device in an automobile’s intermittent wiper system is based on an RC time constant and utilizes a 0.500-μF capacitor

and a variable resistor. Over what range must R be made to vary to achieve time constants from 2.00 to 15.0 s?

Computers and Technology

1 answer:

Tatiana [17]2 years ago

5 0

Answer:

4000k-ohm to 10,000k-ohm

Explanation:

As we know that time constant for an RC circuit is t=RC

Putting the values of t we can get the range of varaiable resistor as;

t=RC

Putting t=2 we get the first value of the range for the variable resistor

2=R*0.500*10^-6

R=2/(0.500*10^-6)

R=4*10^6

R=4000k-ohm

Now putting t=5 we get the final value for the range of variable resistor

t=RC

5=R*0.500*10^-6

R=5/(0.500*10^-6)

R=10*10^6

R=10,000k-ohm

So variable resistance must be made to vary in the range from 4000k-ohm to 10,000k-ohm

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a. 1365 ways

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c. Probability = 0.5904

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$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

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$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

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Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

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Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

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Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

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