Problem: 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? 

FREE Expert Solution

Equivalent capacitance for two capacitors:

Ceq=C1C2C1+C2

The time constant for RC circuits:

τ=RC

Discharging a capacitor in RC circuits:

Q=Q0(1-e-t/τ)

Ceq = (C1C2)/(C1 + C2) = (3C * 5C)/(3C + 5C) = (15/8)C = (15/8)0.13 = 0.24375 F

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Problem Details

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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