229.0 s2 the capacitor has capacitance of C 12.0 uF and the battery maintains th

Question

229.0 s2 the capacitor has capacitance of C 12.0 uF and the battery maintains the emf of & 50.0 V. The switch is closed at time t = 0.0 s and the caDacitor begins to charge. What is the time constant for this circuit? Submt Answer Tries 0/8 What is the charge on the capacitor after the switch has been closed for t = 3.13×103 s? Subent Answer Tries 0/8 What is the current through the circuit after the switch has been closed for t = 3.13 x 10,? Submit Answer Tries 0/8 what is the voltaqe across the capacitor after the switch has been closed for t = 3.13x10,? Type here to

Explanation / Answer

At the moment the switch is closed, the capacitor is modeled as a short circuit.

I(0) = E/R = 50/229 = 0.21834 A

Time constant tau = RC = 229*12x10^-6= 2.748 x 10^-3 s <------------------ 1)

We know voltage across C is given by:

Vc = E[1-e^(-t/RC)]

So:

Vc(t = 3.13x10-3) = 50[1-e^(-(3.13x10^-3)/(2.748x10^-3))]

=> Vc = 34 V

Now,

Qc = C *Vc

Qc(t = 3.13×10-3) = 12x10^-6 *34 = 4.079 x 10^-4 C <------------------ 2)

I(t = 3.13×10-3) = I(0) e^(-t/RC)

=0.21834e^(-(3.13x10^-3)/(2.748x10^-3)) = 0.0699 A <------------------ 3)

Vc(t = 3.13×10-3) = 34V (already calculated in part2)<------------------ 4)

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