# RC circuit discharge while being connected to a battery?

I am facing a conceptual problem in analyzing the following situation which was part of a physics exam;

A series RC circuit with $$R = 200k \Omega$$ and $$C = 10 \mu F$$ is connected to a 5V battery and completely charged after which the battery is exchanged with another battery of opposite polarity, consequently what will be the time after which the voltage across the capacitor become zero?

My approach was to consider the initial Voltage across the capacitor to be 5V, after insertion of the reverse polarity battery, the potential difference is double the initial value i.e 10V, If i were to consider the discharge equation the time $$t = 2 ln2 = 1.38 s$$, but this does not seem to be accurate and i still cant figure the correct equation and logic behind this problem since most standard discharge equations do not involve the presence of an external emf source.

I would also appreciate if someone could point towards relevant sources or similar problems, since my search did not bring any results .

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• Okay, the problem is that you're treating the capacitor as a battery, but it is not. Capacitors work different. The current through them can be written as $C dV/dt$. You should solve it as any linear circuit: appliying Kirchoff's laws. Since $dV/dt$ is a derivative, you will have a differential equation for the voltage across the capacitor. – FGSUZ Mar 8 at 18:20
• @FGSUZ , thank you for pointing that out . now ive got : c dv/dt - IR + 5V = 0 , How do i substitute for the current I ? – Anamika Ghosh Mar 8 at 18:42
• i may have a made a mistake in writing c dv/dt , could you suggest the correction ? – Anamika Ghosh Mar 8 at 18:50
• hint: $cdv/dt$ is the current supplied by the capacitor when its voltage is $v(t)$ – hyportnex Mar 8 at 18:54
• Hi Anamika, if you google wiki RC circuit you will see they show the calculus and the final equation. If you put 10V in the final eqn you should get the right time answer. – PhysicsDave Mar 8 at 19:30

A capacitor charged by a 5V battery becomes a momentary 5V battery, when the polarity change occurs with a fresh 5V battery, the potential across the capacitor is the same as the potential across the battery but in reverse polarity. Hence the capacitor discharges just like how it will without a fresh battery connected, so the discharge time will be calculated normally.

A fully charged capacitor discharges to 63% of its voltage after one time period. After 5 time periods, a capacitor discharges up to near 0% of all the voltage that it once had. Therefore, it is safe to say that the time it takes for a capacitor to discharge is 5 time constants.

To calculate the time constant of a capacitor, the formula is τ=RC. This value yields the time (in seconds) that it takes a capacitor to discharge to 63% of the voltage that is charging it up. After 5 time constants, the capacitor will discharge to almost 0% of all its voltage.

$$τ_t=5RC=5(2)=10 s$$

• i speculated this to be the answer although my exam was multiple choice and the options were 1.96 s, 0.08s ,5.3 s and 1.21 s – Anamika Ghosh Mar 8 at 20:08
• That's surprising. – TechDroid Mar 8 at 20:11
• Are you certain the fresh battery is 5V? – TechDroid Mar 8 at 20:12
• i think the battery was not mentioned to be explicitly 5 V , now that i have rechecked , it simply says "opposite polarity" so i assumed 5 V, but how would that change the answer , could you please explain that – Anamika Ghosh Mar 8 at 20:25
• is it a trick question because the cap only has to get to 0 volts not all the way to -5? – PhysicsDave Mar 8 at 22:40

If I understood the problem correctly, we've got a battery of +5V connected to a $$200 \Omega$$ resistor, and $$10\mu F$$ capacitor afterwards. Then, we suddenly change the battery for a $$-5V$$ one, right? If so, the scheme would be kind of like this:

Stage 1 (before the battery switch)

So, we suppose that, initially, the capacitor has reached a stationary value of $$+5V$$. It has been being charged from wherever to $$+5V$$, just like the battery itself. Of course it is an exponential. The capacitor tends to $$+5V$$ asymptotically, but we can approximate that it is $$+5V$$, ideally.

At this point, the capacitor is non-conducting anymore, so we just see a $$+5V$$ battery connected to a resistor. No current flows through the capacitor, so it behaves like an open circuit. As a consequence, it is like an unconnected branch, no voltage drop across the resistance, so the voltage drop across the capacitor is the same as the battery. $$v=5V$$.

That's what we see until the blue line.

Stage 2 (battery switch)

We press the conmutator and now the battery suddenly changes to $$-5V$$. This keeps being an R-C circuit, so we must solve again for Kirchhoff laws:

$$V_{bat}=V_{R}+V_C$$

It basically says that "the voltage drop through the elements is the same as the voltage inyected by the battery. Equivalently, "sum of voltages = 0". The thing is that we can write:

$$V_{C}=V_{bat}-V_R$$

And we do know the voltage drop across a resistance. It is given by Ohm's law:

$$V_C=V_{bat}-I\cdot R$$.

The voltage of the battery is fixed ($$-5V$$). And what is $$I$$? Well, it is a series circuit. The intensity must be the same across $$R$$ and $$C$$. And we know that the current through the capacitor is $$I_C=C \frac{dV_C}{dt}$$. Thus, we have:

$$V_C=-5V - RC \frac{dV_C}{dt}$$

Check that $$V_C$$, the unknown, is both alone and inside a derivative, so this is a differential equation.

Fortunately, it is separable:

$$\frac{dV}{dt}=\frac{-1}{RC} (V+5V) \qquad \Rightarrow \qquad \frac{dV}{V+5V}=\frac{-dt}{RC}$$

This is easy:

$$\left[\ln (V+5V) \right]_{t=0}^{t}=\frac{-\Delta t}{RC}$$

$$\ln (V+5V) -\ln (+5V+5V) =\frac{-\Delta t}{RC}$$

$$\ln (V+5V) -\ln (10V) =\frac{-\Delta t}{RC}$$

$$\ln \left(\frac{V+5V}{10V}\right) =\frac{-\Delta t}{RC}$$

$$\left(\frac{V+5V}{10V}\right) =e^{-\Delta t/RC}$$

$${V =-5V+10V\cdot e^{-\Delta t/RC}}$$

This is coherent with the scheme. Now, I guess you can get your answer directly.

• is there a problem with this formula a t=0? Should the answer be 5V or 0? – PhysicsDave Mar 8 at 22:44
• According to what I said, if the battery is initially at $5V$ and we suppose that the capacitor has reached stationary regime, so that $V_C=V_{bat}$, then yes, isn't it? – FGSUZ Mar 9 at 0:12
• So what I have to do according to the question is plug V= 0 , I've tried this before but then it isn't giving an answer that matches with the exam . Since t =0 and t =infinity gives V =0 across the cap – Anamika Ghosh Mar 9 at 4:16
• Oops! You're right. Now I get what you mean. The problem is that I introduced wrong limits of integration. Because I'm used to the ones I've put there. I'm correcting this, sorry. – FGSUZ Mar 9 at 11:48
• I had done it quickly. Now it is corrected. Check that now it makes sense, and it reproduces the graph shown in the picture. – FGSUZ Mar 9 at 11:53

Let's solve this question. What you're actually doing is putting a pulse voltage source to the RC-circuit with.

Well, when we have a series RC-circuit we can use Laplace transform to analyse it in detail. Using Faraday's law we can write:

$$\text{v}_\text{s}\left(t\right)=\text{v}_\text{R}\left(t\right)+\text{v}_\text{C}\left(t\right)\tag1$$

Using the relations of the voltage and current in a resitor and a capacitor we can rewrite equation $$(1)$$ as follows:

$$\text{v}_\text{s}'\left(t\right)=\text{i}_\text{R}'\left(t\right)\cdot\text{R}+\text{i}_\text{C}\left(t\right)\cdot\frac{1}{\text{C}}\tag2$$

Because it is a series circuit we know that the input current, $$\text{i}_\text{in}\left(t\right)$$, is the same as the current trough the resistor and the capacitor so we can write:

$$\text{v}_\text{s}'\left(t\right)=\text{i}_\text{in}'\left(t\right)\cdot\text{R}+\text{i}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}\tag3$$

Using the Laplace transform and assuming that the intial conditons are equal to $$0$$ we can write for equation $$(3)$$:

$$\text{s}\cdot\text{V}_\text{s}\left(\text{s}\right)=\text{s}\cdot\text{I}_\text{in}\left(\text{s}\right)\cdot\text{R}+\text{I}_\text{in}\left(\text{s}\right)\cdot\frac{1}{\text{C}}\space\Longleftrightarrow\space\text{I}_\text{in}\left(\text{s}\right)=\frac{\text{s}\cdot\text{V}_\text{s}\left(\text{s}\right)}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag4$$

Writing the supply voltage in the s-domain we get:

$$\text{V}_\text{s}\left(\text{s}\right)=\int_0^{t_1}\hat{\text{u}}\cdot e^{-\text{s}t}\space\text{d}t+\int_{t_1}^\infty\text{u}\cdot e^{-\text{s}t}\space\text{d}t=\frac{1}{\text{s}}\cdot\left(\hat{\text{u}}\cdot\left(1-\exp\left(-\text{s}t_1\right)\right)+\text{u}\cdot\exp\left(-\text{s}t_1\right)\right)\tag5$$

Where $$\hat{\text{u}}$$ is the high (positive) voltage potential (in your case $$+5\space\text{V}$$ and $$\text{u}$$ is the low (negative) voltage potential (in your case $$-5\space\text{V}$$.

So, for the input current we get:

$$\text{I}_\text{in}\left(\text{s}\right)=\frac{\text{s}}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\cdot\frac{1}{\text{s}}\cdot\left(\hat{\text{u}}\cdot\left(1-\exp\left(-\text{s}t_1\right)\right)+\text{u}\cdot\exp\left(-\text{s}t_1\right)\right)=$$ $$\frac{\hat{\text{u}}\cdot\left(1-\exp\left(-\text{s}t_1\right)\right)+\text{u}\cdot\exp\left(-\text{s}t_1\right)}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag6$$

So, the voltage across the capacitor is given by:

$$\text{V}_\text{c}\left(\text{s}\right)=\frac{1}{\text{s}\cdot\text{C}}\cdot\frac{\hat{\text{u}}\cdot\left(1-\exp\left(-\text{s}t_1\right)\right)+\text{u}\cdot\exp\left(-\text{s}t_1\right)}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag7$$

And for the voltage across the resistor we get:

$$\text{V}_\text{R}\left(\text{s}\right)=\text{R}\cdot\frac{\hat{\text{u}}\cdot\left(1-\exp\left(-\text{s}t_1\right)\right)+\text{u}\cdot\exp\left(-\text{s}t_1\right)}{\text{s}\cdot\text{R}+\frac{1}{\text{C}}}\tag8$$

Using your values: $$\text{R}=200\cdot10^3\space\Omega$$, $$\text{C}=10\cdot10^{-6}\space\text{F}$$, $$\hat{\text{u}}=5\space\text{V}$$, $$\text{u}=-5\space\text{V}$$ and assume a switch time of $$1$$ minute we get (using inverse Laplace transform):

• $$\text{i}_\text{in}\left(t\right)=\frac{\exp\left(-\frac{t}{2}\right)}{40000}\cdot\left(1-\exp\left(30\right)\cdot\theta\left(t-60\right)\right)\tag9$$
• $$\text{v}_\text{C}\left(t\right)=5\cdot\left(1-\exp\left(-\frac{t}{2}\right)\right)+10\cdot\left(\exp\left(30-\frac{t}{2}\right)-1\right)\cdot\theta\left(t-60\right)\tag{10}$$
• $$\text{v}_\text{R}\left(t\right)=5\cdot\exp\left(-\frac{t}{2}\right)\cdot\left(1-2\cdot\exp\left(30\right)\cdot\theta\left(t-60\right)\right)\tag{11}$$

Plotting the solution, using Mathematica gives:

Where the blue curve is the input voltage, the orange curve is the voltage across the resistor, the green curve is the voltage across the capacitor and the red curve is the input current (also the current trough the components).