Respuesta :
Answer:
(A). The capacitance of the circuit is [tex]2.31\times10^{-6}\ F[/tex].
(B). The time constant of the circuit is 7.16 sec.
Explanation:
Given that,
Potential = 12.0 V
Internal resistance = 3.10 MΩ
Time = 4.10 s
Voltage = 3.2 V
(A). We need to calculate the capacitance of the circuit
Using formula of capacitance
[tex]V_{C}=V_{0}(e^{\dfrac{-t}{RC}})[/tex]
Put the value into the formula
[tex]3.2=12.0e^{\dfrac{-4.10}{3.10\times10^{6}C}}[/tex]
[tex]e^{\dfrac{-4.10}{3.10\times10^{6}C}}=0.267[/tex]
[tex]\dfrac{-4.10}{3.10\times10^{6}C}=log 0.267[/tex]
[tex]C=\dfrac{-4.10}{3.10\times10^{6}\times log0.267}[/tex]
[tex]C=0.000002306\ F[/tex]
[tex]C=2.31\times10^{-6}\ F[/tex]
The capacitance of the circuit is [tex]2.31\times10^{-6}\ F[/tex]
(B). We need to calculate the time constant of the circuit
Using formula of time constant
[tex]\tau=R\times C[/tex]
Put the value into the formula
[tex]\tau=3.10\times10^{6}\times2.31\times10^{-6}[/tex]
[tex]\tau=7.16\ sec[/tex]
The time constant of the circuit is 7.16 sec.
Hence, (A). The capacitance of the circuit is [tex]2.31\times10^{-6}\ F[/tex].
(B). The time constant of the circuit is 7.16 sec.
