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nikitadnepr [17]

2 years ago

12

Maximum current problem. If the current on your power supply exceeds 500 mA it can damage the supply. Suppose the supply is set

for 37 V. What is the smallest resistance you can measure?

1 answer:

VikaD [51]2 years ago

8 0

To solve this problem we will apply Ohm's law. The law establishes that the potential difference V that we apply between the ends of a given conductor is proportional to the intensity of the current I flowing through the said conductor. Ohm completed the law by introducing the notion of electrical resistance R. Mathematically it can be described as

$V = IR \rightarrow R = \frac{V}{I}$

Our values are

$I = 500mA = 0.5A$

$V = 37V$

Replacing,

$R = \frac{V}{I}$

$R = \frac{37}{0.5}$

$R = 74 \Omega$

Therefore the smallest resistance you can measure is $74 \Omega$

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In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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