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The microcurrent through the electrode in a delicate circuit
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20 Nov 2017, 11:43
20 Nov 2017, 11:43
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Question Stats:
81% (00:41) correct
18% (00:34) wrong based on 94 sessions
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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. \(3.6 * 10^{(-\frac{8}{3})}\)
Kudos for correct solution.
A. \(3.6 * 10^{(-8000)}\)
B. \(3.6 * 10^{(-24)}\)
C. \(3.6 * 10^{(-11)}\)
D. \(3.6 * 10^{(-5)}\)
E. \(3.6 * 10^{(-\frac{8}{3})}\)
Kudos for correct solution.
Re: The microcurrent through the electrode in a delicate circuit
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14 Jan 2018, 02:39
14 Jan 2018, 02:39
1
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)
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)
Re: The microcurrent through the electrode in a delicate circuit
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25 Jan 2018, 07:08
25 Jan 2018, 07:08
1
Just divide the given equation 3.6*(1e-8) by 1.000 (or 10³) to get 3.6*(1e-11).
-> Answer C is correct
-> Answer C is correct
