A capacitor of capacitance C takes 2 s to reach 63% of its maximum charge when c

Question

A capacitor of capacitance C takes 2 s to reach 63% of its maximum charge when connected in series to a resistance R and a battery of emf. How long does it take (from zero initial charge) for this capacitor to reach 95% of its maximum charge?       a) 7 s       b) 4 s       c) 3 s d) 5 s e) 6 s A capacitor of capacitance C takes 2 s to reach 63% of its maximum charge when connected in series to a resistance R and a battery of emf. How long does it take (from zero initial charge) for this capacitor to reach 95% of its maximum charge?       a) 7 s       b) 4 s       c) 3 s d) 5 s e) 6 s 7 s 4 s 3 s 5 s 6 s       a) 7 s       b) 4 s       c) 3 s d) 5 s e) 6 s

Explanation / Answer

Answer is 6 s......option e)
Q = Qo*(1 - e^(-t/T))

Q/Qo = 1- e(-t/T))

e^(-t/T) = 1 - Q/Qo

e^(-t/T) = 1 - 0.63

-t/T = ln(0.37)

T = -t/ln(0.37)

T = 2.01156 (time constant)

now

Q = Qo*(1 - e^(-t/T))

Q/Qo = 1- e(-t/T))

e^(-t/T) = 1 - Q/Qo

e^(-t/T) = 1 - 0.95

-t/T = ln(0.05)

t = -T*ln(0.05)

t = 6 s

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