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?
Electronic – Time taken to charge the capacitor
capacitor
Related Solutions
As it is not a piecewise continuous system, you'll need to calculate, in two steps:
First: you'll need to calculate the time of charging the capacitor until it reaches
$$ (Vb-Vc)/R = Imax $$
with constant current of Imax. if the current is constant that capacitance does not change this is a simple straight ramp curve upto the point where the current is no longer limited by the constant current.
Second: you can use the equations for power supply without limited current power supply.
Which are the exponential curves from that point onward to the asymptote.
Some of the things you say are conflicting, so I want to be clear about the question I am answering. If this is not the question you intended, then you need to write a better question.
The question is what is the voltage accross C1 over time when it initially starts at 0. You didn't specify a capacitance, so we'll leave that as the variable C.
There are two separate regimes to the C1 voltage over time. The first is when the supply is in current limit mode. In that case, the capacitor is being charged up linearly. When the capacitor reaches 9 V, the supply switches over to constant voltage operation. After that, there is a exponential decay from 9 V to 10 V governed by the RC time constant.
The voltage rise of a capacitor is dV = I dT / C. We know the current (I) is 1 Amp and that dV will be 9 V in the first part of the function. The time to reach 9 V is dT = dV C / I. For example, if C = 470 µF, then the time to charge from 0 to 9 V is 4.2 ms. The capacitor voltage will rise linearly during that time.
From 9V on, the supply will be at a constant voltage of 10 V. This remaining 1 V rise will occur as a exponential according to the time constant RC. Again using 470 µF as example for C, that time constant would be 470 µs. That means, for example, that 470 µs after the capacitor has reached 9 V, it will have gained another 630 mV, which would put it at 9.63 V in absolute terms.
Added to clarify why the crossover point is 9 V:
Work backwards and assume the supply is always putting out 10 V. At what capacitor voltage does that require 1 A or more? Since R1 is 1 Ω, 1 A thru it causes a 1 V drop. If the supply is 10 V and R1 drops 1 V, then there must be 9 V on the capacitor when the current is 1 A. If the capacitor voltage is lower, then the voltage on R1 must be higher, which means the supply has to source more than 1 A. However, we know the supply puts out the lesser of 10 V or 1 A, so delivering more than 1 A is not possible. This means for capacitor voltages below 9 V, the supply voltage will be the capacitor voltage plus 1 V, since 1 V will be accross R1 when the supply is putting out 1 A.
Best 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.