Time required to charge a capacitor

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SUMMARY

The discussion focuses on calculating the time required to charge a capacitor with a capacitance of 28.7 μF to a voltage of 28.9 V using a constant current of 63.2 nA. The relevant equations include q = C * Vb * (1 - e^(-t/RC)) and Vb = IR + Qc. The solution involves determining the charge (Q) on the capacitor plates at 28.9 V and using the constant current to find the time (t) for this charge to accumulate, negating the need for the exponential term in this scenario.

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  • Knowledge of capacitance and charge relationships
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Homework Statement


A capacitor with a capacitance of C = 28.7 μF is slowly charged by a constant current of I = 63.2 nA. How long does it take to charge the capacitor to a voltage of V = 28.9 V?


Homework Equations


q=cvb(1-e^(-t/RC))
Vb=IR+QC


The Attempt at a Solution


I have tried to solve for R by R=I/V but I don't know how to solve for t because Q is not given, and I don't know how to find it. Any ideas? I only got a couple hours left. Any help is greatly appreciated.
 
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The statement that the current is constant says that you are in the linear region all the way to charging the capacitor to 28.9 V. You can calculate how much charge is on the plates when the voltage is 28.9 V and from this the time it takes for this charge to accumulate at a constant rate. The exponential is not needed here.
 

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