Time required to charge a capacitor

[Liketothink]

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.

 

[kuruman]

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.

 

[tomcue]

I would approach this problem by first understanding the concept of charging a capacitor. A capacitor is an electrical component that can store energy in the form of an electric charge. When a capacitor is connected to a power source, it starts to accumulate charge until it reaches its maximum capacity. The time required to charge a capacitor depends on its capacitance, the current flowing through it, and the voltage it needs to reach.

In this problem, we are given the capacitance of the capacitor (C = 28.7 μF), the current flowing through it (I = 63.2 nA), and the voltage it needs to reach (V = 28.9 V). Using the equation Vb=IR+QC, we can solve for the charge stored in the capacitor (Q) by rearranging the equation as Q = Vb - IR.

Now, we can use the equation q=cvb(1-e^(-t/RC)) to solve for the time required to charge the capacitor. This equation relates the charge stored in the capacitor (q) to its capacitance (C), voltage (V), and time (t). Rearranging the equation, we get t = -RC ln(1- Vb/q). Substituting the values we have, we get t = - (28.7 μF) (ln(1-28.9 V/28.7 μF)) = 0.064 seconds.

Therefore, it would take approximately 0.064 seconds to charge the capacitor to a voltage of 28.9 V with a constant current of 63.2 nA. It is important to note that this is an ideal calculation and does not take into account any external factors that may affect the charging time, such as resistance in the circuit or leakage of charge.