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(Solved): A leaky capacitor loses 5% of its charge in 8.5min. What is the effective time constant of th ...
A leaky capacitor loses of its charge in . What is the effective time constant of the system? The time constant, Units What fraction of charge (in \%) will be on the capacitor after ? The charge, Units After what time there will be of the initial charg left on the capacitor? The time, Units
Expert Answer
a) The charge on a capacitor with capacitance C and resistance R decays exponentially with time, following the equation: where Q0 is the initial charge on the capacitor and RC is the time constant of the system.In this case, the capacitor loses 5% of its charge in 8.5 min. We can use this information to find the time constant: Dividing both sides by Q0 and taking the natural logarithm of both sides, we get: Solving for RC, we get: Therefore, the effective time constant of the system is approximately 38.02 min.