# Charging capacitor, with series and parallel resisters

I understand how equations for charging capacitor are derived and what they represent. But here is my problem: if you want to experimentally see how voltage on capacitor changes with time, you have to make the circuit down below, the equations for charging capacitor is still valid when $$R_1$$ is much smaller than Voltmeter's resistor. But this isn't true if resistor $$R_1$$ is quite large compared to $$R_2$$.

My question: What is the equation that describes how voltage on charging capacitor is changing with respect to time, where resistor $$R_1$$ isn't very small compared to $$R_2$$.

I don't want just equation pasted in the answer section, I want the explanation how to derive it, and explanation of steps which have to be taken to get there.

Thank you very much in advance. You can solve this circuit with the same differential equation approach you would for the simpler RC circuit. Just grit your teeth and work through the math. I prefer to work transients problems like this using the Laplace Transform method. The circuit is transformed to the frequency domain, solved, and then inverse transformed back to the time domain.

In your example, the battery voltage V converts to a voltage source with amplitude $$\frac{V}{s}$$. The capacitor converts to an impedance of $$\frac{1}{sC}$$. $$R_1$$ and $$R_2$$ are unchanged. "s" is the Laplace Transform operator. Once you have the Laplace Transform circuit you solve it with all the normal electrical circuit techniques. The capacitor and $$R_2$$ in parallel can be lumped together in an equivalent impedance of, $$\frac{\frac{R_2}{sC}}{R_2+\frac{1}{sC}}$$

From that we can find $$I(s)$$ as, $$I(s)=\frac{\frac{V}{s}}{R_1+\frac{\frac{R_2}{sC}}{R_2+\frac{1}{sC}}}$$

Working through the algebra this becomes, $$I(s)=V\frac{\frac{1}{R_1}(s+\frac{1}{R_2C})}{s^2+s(\frac{1}{R_1C}+\frac{1}{R_2C})}$$

You can work this into a common Laplace Transform pair, or you can reduce it using partial fraction expansion. Either way, you are putting into a form that can be transformed back into time domain. I just used the invLaplace operator in MathCad to get the solution,

$${\mathscr{L}}^{-1}I(s) = i(t) = V\frac{R_1+R_2e^{-\frac{t(R_1+R_2)}{CR_1R_2}}}{R_1(R_1+R_2)}$$

If you look at the above it makes sense. Assuming the capacitor is completely discharged when the switch is thrown to connect the battery, the initial current (t=0+) is only restricted by $$R_1$$ and will be amplitude $$V/R_1$$. After the capacitor is substantially charged, its voltage will equal the voltage dropped across $$R_2$$ and the current from the battery will be $$V/(R_1+R_2)$$.

If are interested in the transient voltage across the capacitor, one approach is to take the $$I(s)$$ we found above and multiply it times the impedance of the $$C$$ and $$R_2$$ parallel combination,

$$I(s)(\frac{\frac{R_2}{sC}}{R_2+\frac{1}{sC}})$$

Then carefully work through the math to inverse transform back to the time domain as described above. Doing this in MathCad I get the following for the voltage across the capacitor, $$V_C(t)$$,

$$V_C(t)=-\frac{VR_2[e^\frac{-t(R_1+R_2)}{CR_1R_2}-1]}{R_1+R_2}$$

From this you can see that initially (t=0) the voltage will be zero, and after the capacitor is substantially charged it will be $$\frac{VR_2}{R_1+R_2}$$ which is just voltage division ignoring the fully charged capacitor.

To put some numbers to it, let $$R_1=500k\Omega$$, $$R_2=5M\Omega$$, $$V=50V$$, and $$C=100μF$$ and we find,

$$i(t) = \frac{e^{\frac{-11t}{500}}}{11,000}+\frac{1}{110,000}$$ and $$V_C(t)=\frac{500}{11}-\frac{500e^{\frac{-11t}{500}}}{11}$$

From this you can see that the voltage across the capacitor is initially zero, and eventually 45.455 V.

Here are a couple more examples of solving electrical transients circuits using Laplace:

https://electronics.stackexchange.com/questions/525974/equation-20-16-proof/527788#527788

https://electronics.stackexchange.com/questions/531287/closed-form-solution-for-capacitor-charged-with-current-source-and-discharged-th

To start we find the voltage drop across resistor $$R_2$$ with no capacitor present, this will give us the voltage that the capacitor will charge to when reaching steady state. The voltage drop across $$R_2$$ is

$$V_0=V_T\frac{R_2}{R_1+R_2},$$

where $$V_T$$ is the voltage of the battery. Then we write down the rate equation for the voltage across a charging capacitor, this can be written

$$\dot{V}(t)=A_1\left(V_T-V(t)\right)-A_2V(t),$$

the first positive term is due to the capacitor charging through the resistor $$R_1$$, which is proportional to the difference of the maximum voltage and the acquired voltage during charging. The second negative term is due to the capacitor simultaneously discharging through resistor $$R_2$$. Solving the rate equation gives

$$V(t)=\frac{A_1V_T}{A_1+A_2}\left(1-e^{-(A_1+A_2)t}\right),$$

the coefficients $$A_1$$ and $$A_2$$ are the normal time constants for a capacitor in series with a resistor $$A_{1,2}=1/R_{1,2}C$$. Substituting this into the above and the expression for $$V_0$$ gives

$$V(t)=V_0\left(1-e^{-\frac{R_1+R_2}{R_1R_2C}t}\right).$$

For $$R_2\gg R_1$$, it can be seen that we obtain the standard equation for the voltage across a capacitor in series with a resistance $$R_1$$ as a function of the time.

• @relayman357 That’s correct, I’ve edited the answer. Thanks. May 17, 2021 at 22:54
• Hi @jamie1989, what other circuits will your method work to solve? Just those with one storage element (e.g. 1 cap or 1 inductor)? Just wondering if this is only useful in few cases or more widely. May 19, 2021 at 20:24