Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge
The time that it would take the capacitor to reach 99.99% will be:
"6 × 10⁻⁷ second".
According to the question,
Resistance, R = 12 Ω
Capacitor, C = 500 μf
We know that,
→ T = RC
= 12 × 500 × 10⁻⁶
= 0.006 sec
The max charge be:
→ Q = [tex]\frac{99.99}{100}[/tex] × Qo
= 0.9999 Qo
We know the equation,
→ Q = Qo . exp(-[tex]\frac{t}{RC}[/tex])
By substituting the values,
0.9999 = exp(-[tex]\frac{t}{0.006}[/tex])
By taking "log" both sides, we get
ln(0.9999Qo) = ln(exp(-[tex]\frac{t}{0.006}[/tex]))
-1 × 10⁻⁴ = -[tex]\frac{t}{0.006}[/tex]
t = 6 × 10⁻⁷ second
Thus the above answer is correct.
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