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The microcurrent through the electrode in a delicate circuit

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Bunuel
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The microcurrent through the electrode in a delicate circuit [#permalink] New post   20 Nov 2017, 11:43
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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. \(3.6 * 10^{(-8000)}\)

B. \(3.6 * 10^{(-24)}\)

C. \(3.6 * 10^{(-11)}\)

D. \(3.6 * 10^{(-5)}\)

E. \(3.6 * 10^{(-\frac{8}{3})}\)


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amorphous
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Re: The microcurrent through the electrode in a delicate circuit [#permalink] New post   14 Jan 2018, 02:39
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The microcurrent is usually held at 3.6*10^{(-8)} however this has to be reduced by 1000 times or 10^3
If we multiply 3.6*10^{(-8)} by 10^3 we get 3.6*10^{(-5)} which is bigger than the original number not smaller
so we have to multiply 3.6*10^{(-8)} by 10^{(-3)}
(c)
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Matteo91
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Re: The microcurrent through the electrode in a delicate circuit [#permalink] New post   25 Jan 2018, 07:08
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Just divide the given equation 3.6*(1e-8) by 1.000 (or 10³) to get 3.6*(1e-11).
-> Answer C is correct
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The microcurrent through the electrode in a delicate circuit [#permalink] New post   21 Nov 2018, 08:50
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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. \(3.6*10^{-8000}\)

B. \(3.6*10^{-24}\)

C. \(3.6*10^{-11}\)

D. \(3.6*10^{-5}\)

E. \(3.6*10^{\frac{-8}{3}}\)
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Re: The microcurrent through the electrode in a delicate circuit [#permalink] New post   23 Nov 2018, 13:43
Carcass wrote:
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. \(3.6*10^{-8000}\)

B. \(3.6*10^{-24}\)

C. \(3.6*10^{-11}\)

D. \(3.6*10^{-5}\)

E. \(3.6*10^{\frac{-8}{3}}\)


1,000 times smaller means we divide it by 1,000 which is basically 10^3

So (3.6 * 10^-8)/(10^3) = 3.6 * 10^(-8 -3) = 3.6 * 10^-11

Answer choice C
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Re: The microcurrent through the electrode in a delicate circuit [#permalink] New post   27 Nov 2018, 06:52
thanks
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GreenlightTestPrep
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Re: The microcurrent through the electrode in a delicate circuit [#permalink] New post   18 Nov 2019, 09:01
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Bunuel wrote:
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. \(3.6 * 10^{(-8000)}\)

B. \(3.6 * 10^{(-24)}\)

C. \(3.6 * 10^{(-11)}\)

D. \(3.6 * 10^{(-5)}\)

E. \(3.6 * 10^{(-\frac{8}{3})}\)


Key concept #1: If X is 1000 times smaller than Y, we can write: X = Y/1000

Key concept #2: \(\frac{ab}{c}=(a)(\frac{b}{c})\)


So, the measure of the resulting current = \(\frac{3.6*10^{(-8)}}{1000}\)

Rewrite as: \(\frac{3.6*10^{(-8)}}{10^3}\)

Use Concept #2 to rewrite as: \((3.6)\frac{10^{(-8)}}{10^3}\)

Apply Quotient Law for exponents to get: \((3.6)(10^{(-8-3)})\)

Simplify: \((3.6)(10^{(-11)})\)

Answer: C

Cheers,
Brent
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Mridu2024
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Re: The microcurrent through the electrode in a delicate circuit [#permalink] New post   23 Jun 2024, 23:55
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Given Current as constant as 3.6 * 10 ^-8 , if the current is 1000 amp less then we have a value which is 1000 times smaller than this

so dividing 3.6 * 10 ^-8 we get 3.6 *10^-11
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Re: The microcurrent through the electrode in a delicate circuit [#permalink]
23 Jun 2024, 23:55
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