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# Two 4.4 μF capacitors, two 2.5 kΩ resistors, and a 11.4 V source are connected in series.?

Starting from the uncharged state, how long does it take for the current to drop from its initial value to 1.50 mA ?

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- billrussell42Lv 710 months agoFavorite Answer
two 4.4 in series are equal to 2.2

two 2.5k in series are equal to 5

time constant = 2.2µ x 5k = 11 ms

voltage across 5k is 1.5•5 = 7.5 volts

voltage across cap = 11.4–7.5 = 3.9 volts

v = v₀[1–e^(–t/τ)]

3.9 = 11.4[1–e^(–t/11m)]

1–e^(–t/11m) = 0.3421

e^(–t/11m) = 0.6579

–t/11m = ln 0.6579 = –0.4187

t = 4.61 ms

v = v₀[1–e^(–t/τ)]

v₀ is the battery voltage

i is the current after time t

R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant

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