Question

1A)

A circuit consists of a 12.0 *V* battery, a 100
*k*Ω resistor, a 20.0 *μ**F* capacitor in
series with a switch which is initially in the open position. The
capacitor is initially uncharged. Calculate the charge on the
capacitor 6.00 seconds after the switch is closed. Calculate the
current through the resistor 6.00 seconds after the switch is
closed.

1B)

A 20 *μ**F* capacitor has previously charged up to
contain a total charge of *Q*=100 *μ**C* on
it. The capacitor is then discharged by connecting it directly
across a 100−*k*Ω resistor. At what point in time after the
resistor is connected will the capacitor have 13.5
*μ**C* of charge remaining on it?

Answer #1

1A)

**Given**

V = 12.0 V

R = 100 x 10^{3} ohm

C = 20.0 x 10^{-6} F

t = 6.00 s

**Solution**

Q_{max} = CV = 20.0 x
10^{-6} x 12.0

Q_{max} = 240 x 10^{-6}
C

I_{max} = V/R = 12/100 x
10^{3} = 1.2 x 10^{-4} A

Time constant

τ = RC = 100 x 10^{3} x 20.0 x
10^{-6}

τ = 2 s

Q_{t} = Q_{max}
[1-e^{-t/}^{τ}]

Q_{6} = 240 x 10^{-6}
[1-e^{-6/2}]

Q_{6} = 2.28 x 10^{-4}
C

I_{t} = I_{max}
e^{-t/}^{τ}

I_{6} = 1.2 x 10^{-4} x
e^{-3}

I_{6} = 5.97 x 10^{-6}
A

1B)

**Given**

C= 20 x10^{-6} F

R = 100 x 10^{3} ohm

Q_{o} = 100 x 10^{-6}
C

Q = 13.5 x 10^{-6} C

**Solution**

Q_{t} =
Q_{o} [e^{-t/τ}]

13.5 x 10^{-6} = 100 x
10^{-6} [e^{-t/2}]

0.135 = e^{-t/2}

e^{t/2} = 1/0.135 = 7.407

t/2 = ln(7.407)

t = 4.00 s

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