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4vir4ik [10]
3 years ago
11

1. Describe an application where a series circuit might work better than a parallel circuit

Computers and Technology
2 answers:
Solnce55 [7]3 years ago
8 0

If you have a light bulb and you want to be able to switch it on and
off, then you must connect the switch and the light bulb in series
across the battery.

If you connect the switch and light bulb in parallel across the battery,
then the light bulb will shine all the time, and PLUS ... when you flip the
switch to ' ON ', sparks will shoot out of the switch, the wires will get hot
and smoking, and the battery will instantly empty itself. 

So this is an application where the series circuit might work better than
the parallel one.


Len [333]3 years ago
7 0
Another example could be Christmas tree lights. You have sometimes hundreds of lights wired in series. If they were in parallel the wire would then have to go to each light meaning potentially hundreds of different wires
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The Pentium 4 Prescott processor, released in 2004, has a clock rate of 3.6 GHz and voltage of 1.25 V. Assume that, on average,
Zarrin [17]

Answer:

a) C = 3.2 x 10^-8 F, C = 2.9 x 10^-8 F

b) 10%, 42.86%, 0.11, 0.75

c) 0.849 volts to maintain the same leakage current in the Pentium 4 Prescott Processor. 0.783 Volts to maintain the same leakage current in the Core i5 Ivy Bridge Processor.

Explanation:

<u>For the Pentium 4 Prescott processor:</u>

Clock rate (f) = 3.6 GHz

Voltage (V) = 1.25 V

Static Power Consumption = 10 W

Dynamic Power Consumption = 90 W

<u>For the Core i5 Ivy Bridge processor:</u>

Clock rate (f) = 3.4 GHz

Voltage (V) = 0.9 V

Static Power Consumption (SP) = 30 W

Dynamic Power Consumption (DP) = 40 W

(a) To find the average capacitive loads, use the formula:

DP = (1/2) x C x V² x f

Rearranging:

C = (2 x DP)/(V² x f)

For the Pentium 4 Prescott processor:

C = (2 x 90) / [(1.25)² x 3.6 x 10^9]

C = 3.2 x 10^-8 F

For the Core i5 Ivy Bridge processor:

C = (2 x 40) / [(0.9)² x 3.4 x 10^9]

C = 2.9 x 10^-8 F

(b) Percentage of the total dissipated power comprised by the static power can  be found by the formula:

<u>[Static Power/(Static Power + Dynamic Power)] x 100</u>

For the Pentium 4 Prescott Processor:

[10/(90+10)] x 100  = 10%

For the Core i5 Ivy Bridge Processor:

[30/30+40] x 100 = 42.86 %

Ratio of static power to dynamic power for each technology can be computed as:

Pentium 4 Prescott: 10/90 = 0.11

Core i5 Ivy Bridge: 30/40 = 0.75

(c) The total dissipated power can be calculated by:

P = DP + SP

For Static Power we have the formula:

SP = VI where I is the leakage current.

I = SP/V

For Pentium 4 Prescott: I = 10/1.25 = 8A

For Core i5 Ivy Bridge: I = 30/0.9 = 3.33 A

The total disspited power is reduced by 10%, meaning the ratio of new power to old power must be 90% = 0.9. So,

New Power/Old Power = 0.9

For Pentium 4 Prescott:

DP + SP / 100 = 0.9

VI + CV²F = 0.9 x 100

V(8) + (3.2 x 10^-8)V²(3.6 x 10^9) = 90

8V + 115.2V² - 90 = 0

By solving the quadratic equation, we get the positive answer as 0.849. This means that the voltage should be reduced to 0.849 volts to maintain the same leakage current in the Pentium 4 Prescott Processor.

For Core i5 Ivy Bridge:

DP + SP / 70 = 0.9

V(3.33) + (2.9 x 10^-8)V²(3.4 x 10^9) = 70 x 0.9

3.33V +98.6V² -63 = 0

By solving the quadratic equation we get the positive answer as 0.783. This means that the voltage should be reduced to 0.783 Volts to maintain the same leakage current in the Core i5 Ivy Bridge Processor.

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