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Vinvika [58]
3 years ago
8

In an electrical circuit, the current is related to the power and resistance according to the rule I=pR−−√, where I represents t

he current (measured in amps), P represents power (measured in watts), and R represents resistance (measured in ohms). What is the electrical power drawn by a device that has a resistance of 15 ohms with a current of 4 amps?
Mathematics
1 answer:
Anna11 [10]3 years ago
5 0

Answer:

P=0.27\,\,watts

Step-by-step explanation:

Consider the formula: I=PR

Here, I represents the current (measured in amps), P represents power (measured in watts), and R represents resistance (measured in ohms).

Resistance = 15 ohms

Current = 4 amps

To find electrical power drawn by a device, put R=15\,,\,I=4

So,

4=15P\\P=\frac{4}{15}=0.27\,\,watts

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Here's a method that takes advantage of the values of these particular numbers.

... 76 = 80 - 4

so

... 76 × 327 = (80 -4)×327

... = 80×327 - 4×327

Repeated doubling will give us values that are 2, 4, and 8 times 327.

... 2×327 = 327+327 = 654

... 2×654 = 4×327 = 654+654 = 1308 . . . . we'll use this later

... 2×1308 = 8×327 = 2616

We want 80×327, so we can add a zero to the end of this last:

... 80×327 = 26160

Now, we can subtract 4×327 to get 76×327

... 80×327 - 4×327 = 26160 -1308 = 24852 = 76×327

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More conventionally, you would multiply every digit of one number by every digit of the other and add the products according to their respective place values.

327 × 076 = (3×0)×10000 + (3×7 +2×0)×1000 +(3×6 +7×0 +2×7)×100 +(2×6 +7×7)×10 +(7×6)×1

... = 0 +21,000 +3,200 + 610 +42

... = 24,852

Note the pattern of partial products here. This is a method taught to/by practitioners of Vedic mathematics, and can be done in your head. At most, you would write down the partial product sums 21, 32, 61, and 42 to keep from having to carry more than one number in your head at a time.

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