Time taken to charge the capacitor

Which equation can be used to calculate the time taken to charge the capacitor at the given amount of current and voltage at a constant capacitance?

Answer

If you want a "simple" equation, and it seems that you do, you could start with definition of current.

First, let's start with the farad. It is usually expanded as \$F=\frac {As}{V}\$.

Now let's write that with symbols for capacitance, current, voltage and time:

\$C=\frac {It}{U}\$

Since we have constant current and voltage and we need time, we'll divide the equation with current and multiply with voltage so that we can get time.

That gives us \$\frac{UC}{I}=t\$.

If this is just a school problem, then we have a solution.

In real life things will work differently. As the capacitor charges, the voltage on the capacitor will drop resulting in drop of current and the time will therefore be longer.

Here's an example: Let's assume that at the beginning, the capacitor is discharged. First we have the voltage on the resistor which is \$U_r=Ri\$. Then we have voltage on the capacitor which is \$U_c=\frac{1}{C} \int {i \mbox{ }dt} \$.

So we know that \$E=Ri+\frac{1}{C} \int {i \mbox{ }dt}\$. To solve this, we need to turn it into differential equation.

\$(E=Ri+\frac{1}{C} \int {i \mbox{ }dt}) / \frac{d}{dt}\$

Since \$E\$ is constant, it will turn into zero. The integration and differentiation will cancel each-other out and we'll get:

\$R\frac{di}{dt}+\frac{i}{C}=0\$ Next we divide everything with \$R\$ and get
\$\frac{di}{dt}+\frac{i}{RC}=0\$

After that we move the \$\frac{1}{RC}i\$ to the other side and multiply everything with \$dt\$ and divide everything with \$i\$ and we get:

\$\frac{di}{i}=-\frac{1}{RC}dt\$

Now we integrate everything and get \$\int {\frac{di}{i}} = -\int {\frac{1}{RC}dt}\$ As a result, we get:

\$\ln{i}=-\frac{t}{RC}+C_1\$

Now to get rid of the logarithm, we raise everything to \$e\$
\$i=C_1 e^{-\frac{t}{RC}}\$

Now we have the general solution and we need to determine the constants. So first we look at what's happening when the time is equal to zero:

\$i=C_1 e^{-\frac{0}{RC}} = C_1\$.

We also know that the initial current is \$i_{(0)}=\frac{E}{R}\$. From that we can determine that \$C_1=\frac{E}{R}\$.

The complete equation for the current is:

\$i_{(t)}=\frac{E}{R} e^{-\frac{t}{RC}}\$

This is a classical capacitor charging equation and it is available on many sources on the Internet.

The \$RC\$ is also called the time constant, so \$\tau=RC\$. It is usually considered that five time constants are enough to charge a capacitor.