Equivalent capacitance for two capacitors:
The time constant for RC circuits:
Discharging a capacitor in RC circuits:
Ceq = (C1C2)/(C1 + C2) = (3C * 5C)/(3C + 5C) = (15/8)C = (15/8)0.13 = 0.24375 F
Two capacitors of capacitance 3C and 5C (where C = 0.13 F) are connected in series with a resistor of resistance R Randomized Variables R=5.5 Ω
How long will it take the amount of charge in the circuit to drop by 75% in seconds?
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