Answer:
R min = 28.173 ohm
R max = 1.55 × [tex]10^{4}[/tex] ohm
Explanation:
given data
capacitor = 0.227 μF
charged to 5.03 V
potential difference across the plates = 0.833 V
handled effectively = 11.5 μs to 6.33 ms
solution
we know that resistance range of the resistor is express as
V(t) = [tex]V_o \times e^{t\RC}[/tex] ...........1
so R will be
R = [tex]\frac{t}{C\times ln(\frac{V_o}{V})}[/tex] ....................2
put here value
so for t min 11.5 μs
R = [tex]\frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}[/tex]
R min = 28.173 ohm
and
for t max 6.33 ms
R max = [tex]\frac{6.33}{11.5} \times 28.173[/tex]
R max = 1.55 × [tex]10^{4}[/tex] ohm