Respuesta :
Answer:
The time constant is 1.049.
Explanation:
Given that,
Charge [tex]q{t}= 0.65 q_{0}[/tex]
We need to calculate the time constant
Using expression for charging in a RC circuit
[tex]q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}][/tex]
Where, [tex]\dfrac{t}{RC}[/tex] = time constant
Put the value into the formula
[tex]0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}][/tex]
[tex]1-e^{-(\dfrac{t}{RC})}=0.65[/tex]
[tex]e^{-(\dfrac{t}{RC})}=0.35[/tex]
[tex]-\dfrac{t}{RC}=ln (0.35)[/tex]
[tex]-\dfrac{t}{RC}=-1.049[/tex]
[tex]\dfrac{t}{RC}=1.049[/tex]
Hence, The time constant is 1.049.
The time constant must elapse 1.049 times
Time constant:
Let the charge on the capacitor be q.
The discharge of the capacitor is a function of time as described below:
[tex]q(t)=q_o[1-e^\frac{t}{RC} ][/tex]
where RC is the time constant of the circuit
and q₀ is the initial charge
It is given that the charge left is:
q(t) = 0.65q₀
[tex]q(t)=q_o[1-e^-\frac{t}{RC} ]\\\\0.65q_o=q_o[1-e^{-\frac{t}{RC}} ]\\\\e^{-\frac{t}{RC}}=0.35\\\\\frac{t}{RC}=1.049[/tex]
The time constant elapses 1.049 times.
Learn more about time constant:
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