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Answer :
To determine the time it takes for a capacitor to discharge to half its initial voltage, we use the formula for capacitor discharge. Given a 6 F capacitor with an initial voltage of 10V and a 10,000 Ω resistor, it takes approximately 41,580 seconds to reach 5V.
To solve this, we need to use the formula for the discharge of a capacitor:
[tex]V(t) = V_{0} *[/tex] [tex]e^{(-t/RC)}[/tex]
where:
- V(t) is the voltage across the capacitor at time t
- V0 is the initial voltage (10V)
- R is the resistance (10,000 Ω)
- C is the capacitance (6 F)
- t is the time in seconds
We need to find the time t when V(t) = 5V:
[tex]5 = 10 *[/tex] [tex]e^{(-t / (10,000 * 6))}[/tex]
[tex]0.5 =[/tex] [tex]e^{(-t / (10,000 * 6))}[/tex]
ln(0.5) = -t / 60,000
-0.693 = -t / 60,000
t = 0.693 * 60,000
t ≈ 41,580 seconds
Therefore, it takes approximately 41,580 seconds for the voltage across the capacitor to reduce to 5V.
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