Answer:
a) 144.000 s
b) and c)Battery voltage and power plots in attached image.
[tex]V=-\frac{0.5}{144000} t + 1.5 V[tex]
[tex]P(t)=-(31.25X10^{-9}) t+0.0135[/tex] where D:{0<t<40} h
d) 1620 J
Explanation:
a) The first answer is a rule of three
[tex]s=\frac{3600s * 40h}{1h} = 144000s[/tex]
b) Using the line equation with initial point (0 seconds, 1.5 V)
[tex]m=\frac{1-1.5}{144000-0} = \frac{-0.5}{144000}[/tex]
where m is the slope.
[tex]V-V_{1}=m(x-x_{1})[/tex]
where V is voltage in V, and t is time in seconds
[tex]V=m(t-t_{1}) + V_{1}[/tex] and using P and m.
[tex]V=-\frac{0.5}{144000} t + 1.5 V[tex]
c) Using the equation V
POWER IS DEFINED AS:
[tex] P(t) = v(t) * i(t) [tex]
so.
[tex] P(t) = 9mA * (-\frac{0.5}{144000} t + 1.5) [tex]
[tex]P(t) = - (31.25X10^{-9}) t + 0.0135[/tex]
d) Having a count that.
[tex]E = \int\limits^{144000}_{0} {P(t)} \, dt = \int\limits^{144000}_{0} {v(t)*i(t)} \, dt[/tex]
[tex]E = \int\limits^{144000}_{0} {-\frac{0.5}{144000} t + 1.5*0.009} \, dt = 1620 J[/tex]
