Respuesta :
A) Charge supplied: 1512 C
The battery in this problem supplies a charge of:
[tex]Q=0.42 A \cdot h[/tex]
where the charge is written in the form:
[tex]Q=I \cdot t[/tex]
where I is the current and t is the time elapsed. We have:
[tex]I=0.42 A[/tex] is the current
[tex]t=1 h=3600 s[/tex] is the time elapsed
Substituting in the equation, we find
[tex]Q=(0.42 A)(3600 s)=1512 C[/tex]
B) Expected lifetime: [tex]2.7\cdot 10^8 s[/tex]
The average current produced by the pacemaker is
[tex]I=5.6\mu A=5.6\cdot 10^{-6} A[/tex]
While the total charge available is
Q = 1512 C
Therefore, the expected lifetime is
[tex]t=\frac{Q}{I}=\frac{1512 C}{5.6\cdot 10^{-6} A}=2.7\cdot 10^8 s[/tex]
which corresponds approximately to 8.56 years.
Current is the flow of electrons. The amount of charge that will be supplied by this battery will be 1,512 C.
What is current?
Current can be defined as the rate at which electron-deficient atoms or can say electrons flow. It is represented by 'I'. And the SI unit of current is Ampere which is equal to one column of electrical charge.
Given to us
Current, I₀ = 0.42A⋅h
Current produced by the pacemaker, I₁ = 5.6μA = 5.6 x 10⁻⁶ A
A.) The coulombs of charge the battery can supply be found using the relation,
[tex]Q = I \times t\\\\Q = 0.42 \times 3600\\\\Q = 1,512\rm\ C[/tex]
Thus, the amount of charge that will be supplied by this battery will be 1,512 C.
B.) As we have already discussed about charge using the same relationship,
[tex]Q = I \times t[/tex]
We know the charge produced by the battery, also the current produced by the pacemaker, therefore,
[tex]1512 = 5.6 \times 10^{-6} \times t\\\\t= 2.7 \times 10^{-4}\rm\ sec[/tex]
Hence, the expected life of the device is 2.7x10⁻⁴ sec.
Learn more about Current:
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