Consider the following configuration of solar photovoltaic arrays consisting of crystalline silicon solar cells. There are two subsystems connected in parallel, each one containing two cells. In order for the system to function, at least one of the two parallel subsystems must work. Within each subsystem, the two cells are connected in series, so a subsystem will work only if all cells in the subsystem work. Consider a particular lifetime value t0, and suppose we want to determine the probability that the system lifetime exceeds t0. Let Ai denote the event that the lifetime of cell i exceeds t0 (i = 1, 2, , 4). We assume that the Ai's are independent events (whether any particular cell lasts more than t0 hours has no bearing on whether or not any other cell does) and that P(Ai) = 0.6 for every i since the cells are identical. Using P(Ai) = 0.6, the probability that system lifetime exceeds t0 is easily seen to be 0.5904. To what value would 0.6 have to be changed in order to increase the system lifetime reliability from 0.5904 to 0.61? [Hint: Let P(Ai) = p, express system reliability in terms of p, and then let x = p2.] (Round your answer to four decimal places.)
Consider the following configuration of solar photovoltaic arrays consisting of crystalline silicon solar cells. There are two subsystems connected in parallel, each one containing two cells. In order for the system to function, at least one of the two parallel subsystems must work. Within each subsystem, the two cells are connected in series, so a subsystem will work only if all cells in the subsystem work. Consider a particular lifetime value t0, and suppose we want to determine the probability that the system lifetime exceeds t0. Let Ai denote the event that the lifetime of cell i exceeds t0 (i = 1, 2, , 4). We assume that the Ai's are independent events (whether any particular cell lasts more than t0 hours has no bearing on whether or not any other cell does) and that P(Ai) = 0.6 for every i since the cells are identical. Using P(Ai) = 0.6, the probability that system lifetime exceeds t0 is easily seen to be 0.5904. To what value would 0.6 have to be changed in order to increase the system lifetime reliability from 0.5904 to 0.61? [Hint: Let P(Ai) = p, express system reliability in terms of p, and then let x = p2.] (Round your answer to four decimal places.)
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter7: Exponents And Exponential Functions
Section7.1: Multiplication Properties Of Exponents
Problem 1BGP
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Consider the following configuration of solar photovoltaic arrays consisting of crystalline silicon solar cells.
There are two subsystems connected in parallel, each one containing two cells. In order for the system to function , at least one of the two parallel subsystems must work. Within each subsystem, the two cells are connected in series, so a subsystem will work only if all cells in the subsystem work. Consider a particular lifetime value t0, and suppose we want to determine the probability that the system lifetime exceeds t0.
Let Ai denote theevent that the lifetime of cell i exceeds t0
(i = 1, 2, , 4). We assume that the Ai's are independent events (whether any particular cell lasts more than
Let Ai denote the
(i = 1, 2, , 4). We assume that the Ai's are independent events (whether any particular cell lasts more than
t0 hours has no bearing on whether or not any other cell does) and that P(Ai) = 0.6 for every i since the cells are identical. Using P(Ai) = 0.6, the probability that system lifetime exceeds t0 is easily seen to be 0.5904. To what value would 0.6 have to be changed in order to increase the system lifetime reliability from 0.5904 to 0.61? [Hint: Let
P(Ai) = p, express system reliability in terms of p, and then let x = p2.] (Round your answer to four decimal places.)
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Expert Solution
Step 1
From the given circuit we can observe that there are two subsystems connected in parallel.
Each subsystem has 2 cells connected in series.
All the cells are identical and independent.
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