A simple PC circuit consists of a 1MuF capacitor in series with a 3000Ohm resist

Question

A simple PC circuit consists of a 1MuF capacitor in series with a 3000Ohm resistor, a 6V battery, and an open switch Initially, the capacitor is uncharged How long after the switch is dosed will the voltage across the capacitor be 3 8 V? 3 Times 10^9 s 0.003 s 3 s 3 Times 10^-8 s 3 Times 10^9 s A battery' supplies a total charge of 5 0 MuC to a circuit which consists of a series combination of two identical capacitors, each with capacitance C. Determine the charge on either capacitor. 0,50 MuC 1.0 MuC 1.5MuC 2.5MuC 5.0MuC

Explanation / Answer

Voltage after charge capacitor

V = Vs*(1 - e^(-t/RC))

Vs = supply voltage

3.8 = 6*(1 - e^(-t/(1*10^(-6)*3000)))

3.8/6 = 1 - e^(-t*333.33)

e^(-t*333.33) = 0.367

(-t*333.33) = ln(0.367)

(-t*333.33) = -1.0024

t = 0.003 sec (ANSWER of a)

b)

For series combination of capacitor,

charge on all capacitor is same.

so the charge on either capacitor = 5.0 uC

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