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The microcurrent through the electrode in a delicate circuit

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by BTGmoderatorLU » Sun Sep 30, 2018 2:45 pm

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Source: Magoosh

The microcurrent through the electrode in a delicate circuit is usually held constant at 3.6*10^(−8) amps. Because of a defect in another part of the circuit, the current was 1,000 times smaller. What was the current, in amps, caused by this defect?
$$A.\ 3.6\cdot10^{-8000}$$
$$B.\ 3.6\cdot10^{-24}$$
$$C.\ 3.6\cdot10^{-11}$$
$$D.\ 3.6\cdot10^{-5}$$
$$E.\ 3.6\cdot10^{-8/3}$$
The OA is C.
Last edited by BTGmoderatorLU on Mon Oct 01, 2018 2:22 pm, edited 1 time in total.
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Source: — Problem Solving |

by [email protected] » Sun Sep 30, 2018 3:12 pm
Hi All,

We're told that the microcurrent through the electrode in a delicate circuit is usually held constant at (3.6)(10^(-8)) amps and, because of a defect in another part of the circuit, the current was 1,000 times SMALLER. What was the current, in amps, caused by this defect. This question is a 'concept question' based on Scientific Notation, so you don't actually have to do much math if you recognize the concepts involved.

When a number is 10 times GREATER than another number, then we multiply the smaller number by 10 to get the larger number (re: X and 10X).
When a number is 100 times GREATER than another number, then we multiply the smaller number by 100 to get the larger number (re: X and 100X). We can also use Scientific Notation to note the value: 100X = (10^2)(X)).

In simple terms, as we get 'powers of 10' greater, the value of the exponent increases. In that same way, when we get 'powers of 10' smaller, the value of the exponent decreases. In this prompt, we're dealing with a number that is 1000 times SMALLER, so the exponent has to DECREASE by 3. The current begins with an exponent of "-8", so it becomes "-11" when it becomes smaller.

Final Answer: C

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