The capacitor; the RC circuit

What Nour says

1. 1. Capacitor description · charge and capacitance

   Today, the whole chapter: "The capacitor; the RC circuit", Physics, Year 4, Mathematics section.

   We will build this step by step, from the simple capacitor to the full RC circuit.

   Each part has a drawing, a worked example, and a short question to check your understanding.

   Let's start with the capacitor, a small device used in almost every electric circuit.

   This is the brief description of a capacitor (description sommaire d'un condensateur), that is a simple picture of what a capacitor looks like.

   A capacitor has two metal plates, facing each other, separated by a thin insulating layer.

   That insulating layer is called the dielectric, and it does not let charges pass through.

   We label the two plates A and B, and connect wires to each one.

   In circuit diagrams, we draw the capacitor as two short parallel lines.

   Now let's connect this capacitor to a battery, through a switch.

   When the switch closes, the battery pushes charges onto the plates.

   Positive charges pile up on plate A, and negative charges pile up on plate B.

   We call this amount of charge the charge électrique of the capacitor.

   That is the quantity of electricity, measured in coulombs, stored on each plate.

   The two plates always carry the same amount of charge, but with opposite signs.

   This is why we also talk about the capacitance of a capacitor, written with the letter C.

   The capacitance tells us how much charge the capacitor can store for a given voltage.

   A bigger capacitance means the capacitor stores more charge at the same voltage.

   The unit of capacitance is the farad, written F, named after Michael Faraday.

   In practice, one farad is huge, so we usually use microfarads or nanofarads.

   Here is why the capacitor behaves this way: the dielectric blocks direct charge flow.

   So charges cannot cross from plate A to plate B directly.

   Instead, they simply accumulate on each plate, creating an electric field between them.

   This field is what stores the energy, which we will study later in this chapter.

   Now let's look at a worked example, close to a real bac exercise.

   A capacitor stores a charge of 6 microcoulombs when the voltage across it is 3 volts.

   We want to find its capacitance C, using the basic relation q equals C times U.

   We rearrange this formula to isolate C, the capacitance we are looking for.

   Now we substitute the values: 6 microcoulombs divided by 3 volts.

   This gives us 2 microfarads for the capacitance of this capacitor.

   We frame this result: a capacitance of 2 microfarads, C equals 2 microfarads.

   Notice that the capacitance does not depend on the charge or the voltage themselves.

   It only depends on how the capacitor is built, its plates and its dielectric.

   So for a fixed capacitor, the charge is always proportional to the voltage.

   Let's state the key result of this section clearly.

   A capacitor is two conducting plates separated by an insulator, storing opposite charges q and minus q.

   Its capacitance C, in farads, measures how much charge it stores per volt applied.

   What is the SI unit of a capacitor's capacitance?

2. 2. Charge and discharge of a capacitor · charge and current intensity

   Now let's study how a capacitor charges and discharges over time.

   We connect a capacitor, a resistor, and a battery, through a switch, in one loop.

   When we close the switch, current flows and charge builds up on the capacitor.

   This is called charging: the voltage across the capacitor rises from zero toward its final value.

   Here is the voltage u sub C, in volts, plotted against time t, in seconds.

   The curve rises quickly at first, then slows down as it nears its final value.

   If we then open the switch and let the capacitor release its charge, we get discharge.

   The voltage now drops from its high value back down toward zero volts.

   During charging and discharging, charge moves, and that means current flows in the wires.

   The current intensity, written i, tells us how fast the charge changes with time.

   Here is why this makes sense: current is simply charge flow per unit of time.

   When charge increases quickly, the current is large; when it levels off, the current drops.

   During discharge, charge leaves the capacitor, so the current flows in the opposite direction.

   Now let's work through an example, similar to a bac question.

   A capacitor gains a charge of 12 microcoulombs during a time interval of 4 milliseconds.

   We want the average current intensity during this charging interval.

   We substitute: 12 microcoulombs divided by 4 milliseconds.

   This calculation gives us 3 milliamperes for the average current.

   We frame this: the average current intensity is 3 milliamperes, flowing onto the capacitor.

   So the capacitor charges and discharges, never instantly, but over some span of time.

   The current i equals the rate of change of charge q with respect to time.

   During charging, current enters the capacitor; during discharging, it leaves, in the opposite direction.

   During discharge, in which direction does the current flow compared to charging?

3. 3. Relation between charge q and voltage u_C · capacitance of a plane capacitor

   Now let's connect charge, voltage, and capacitance in one clear relation.

   We already met the formula q equals C times U, for the capacitor's charge.

   Here, q is the charge in coulombs, C is the capacitance in farads, and u sub C is the voltage in volts.

   This relation says charge and voltage are always proportional, for a given capacitor.

   If we plot q against u sub C, we get a straight line through the origin.

   The slope of this line is exactly the capacitance C, in farads.

   Here is why: a bigger slope means more charge stored for the same voltage.

   Now let's look at how a capacitor is actually built, as a plane capacitor.

   Two flat plates, each of surface area S, face each other at a small distance d apart.

   The capacitance of this plane capacitor depends on S, d, and the insulator between the plates.

   The formula is C equals epsilon times S divided by d.

   Here epsilon is the permittivity of the insulator, S is the plate area, d is the gap.

   A bigger plate area or a smaller gap both increase the capacitance.

   Now let's solve a worked example, like in a bac exam.

   A plane capacitor has plates of area 0{,}02 square meters, separated by 0{,}1 millimeter of air.

   The permittivity of air is about 8{,}85 times 10 to the minus 12 farads per meter.

   We substitute these values into the formula for C.

   This calculation gives a capacitance close to 1{,}77 nanofarads.

   We frame this result: the capacitance is about 1{,}77 nanofarads.

   Notice how a tiny gap d gives a very small capacitance here.

   So charge and voltage stay proportional through C, and C depends on the capacitor's geometry.

   For a plane capacitor, bigger plates or a thinner gap both raise the capacitance.

   In the formula C equals epsilon times S over d, what happens to C if d gets smaller?

4. 4. Service voltage and breakdown voltage · energy stored in a capacitor

   Let's now look at the limits of a capacitor, and the energy it can store.

   Every real capacitor has a maximum voltage it can safely handle.

   This is called the service voltage, written U sub N, the highest voltage for normal safe use.

   If we go above this limit, the insulator between the plates can suddenly fail.

   That failure point is called the breakdown voltage, that is the voltage where sparks pass through the insulator.

   At breakdown, current suddenly jumps across the dielectric, and the capacitor is often destroyed.

   Here is a simple picture: voltage rises safely, until it reaches the breakdown value.

   Beyond U sub N there is still a safety margin, before reaching U sub c, the breakdown point.

   So we always choose U sub N safely below U sub c, to protect the capacitor.

   Now let's talk about the energy a charged capacitor stores.

   We call this the energy emmagasinée, that is the electric energy kept inside the capacitor's field.

   Here E is in joules, C is in farads, and U sub C is the voltage in volts.

   Here is why this makes sense: more charge or more voltage both mean more stored energy.

   Since q equals C times U sub C, this energy also equals one half q times U sub C.

   A fully discharged capacitor, at zero volts, stores zero energy.

   Now let's solve a worked example, close to a bac exercise.

   A capacitor of 470 microfarads is charged under a voltage of 12 volts.

   We want the energy stored in this capacitor, using E equals one half C U sub C squared.

   We compute 12 squared first, which gives 144.

   Then we multiply: one half times 470 microfarads times 144.

   This calculation gives an energy of about 0{,}0338 joules.

   We frame this result: the stored energy is about 33{,}8 millijoules.

   Notice that the voltage is squared, so doubling U sub C quadruples the stored energy.

   Let's restate the two key ideas from this section.

   The service voltage is the safe limit, always below the breakdown voltage U sub c.

   The stored energy equals one half C times U sub C squared, always measured in joules.

   If we double the voltage U sub C, by what factor does the stored energy increase?

5. 5. Response of an RC circuit to a voltage step

   Now let's study the full RC circuit, and how it reacts to a sudden voltage change.

   We call this the response of an RC circuit to a voltage step (réponse d'un dipôle RC à un échelon de tension).

   That means how a resistor-capacitor circuit reacts when the voltage suddenly jumps from zero to a fixed value.

   We build the circuit: a battery E, a switch K, a resistor R, and a capacitor C, in series.

   The voltmeter measures the voltage u sub C across the capacitor, as it charges.

   Before closing K, the voltage u sub C is zero, and the capacitor is empty.

   At the instant t equals zero, we close the switch: this is the voltage step.

   Here is the curve of u sub C against time t, after closing the switch.

   The voltage u sub C rises from zero and approaches the battery voltage E, but never quite reaches it instantly.

   This dashed line marks the final value, the battery voltage E, in volts.

   The differential equation governing this charging is R C times du sub C over dt plus u sub C equals E.

   The solution to this equation is u sub C equals E times one minus e to the minus t over tau.

   Here tau is the time constant (constante de temps), that is the time that sets how fast the capacitor charges.

   Tau equals R times C, with R in ohms and C in farads, giving tau in seconds.

   Here is why tau matters: after a time tau, u sub C reaches about 63 percent of E.

   After about 5 times tau, the capacitor is considered fully charged, very close to E.

   A small tau means fast charging; a large tau means slow charging.

   Now let's solve a worked example, close to a bac question.

   A circuit has a resistor of 2 kiloohms and a capacitor of 100 microfarads.

   We want the time constant tau, using tau equals R times C.

   We substitute: 2000 ohms times 100 times 10 to the minus 6 farads.

   This calculation gives a time constant of 0{,}2 seconds.

   We frame this result: the time constant tau is 0{,}2 seconds here.

   So the capacitor would reach about 63 percent of E in just 0{,}2 seconds.

   Let's restate the key result of this section.

   When voltage is switched on, u sub C rises following u sub C equals E times one minus e to minus t over tau.

   The time constant tau equals R times C, and it sets how fast charging happens.

   What percentage of the final voltage E is reached after one time constant tau?

6. 6. Discharge of a capacitor through a resistor · effect of R and C on charge and discharge time

   Let's finish this chapter with the discharge of a capacitor through a resistor.

   We take our charged capacitor, remove the battery, and connect it directly to the resistor.

   At the instant t equals zero, the capacitor starts releasing its stored charge through R.

   The voltage u sub C now decreases from its initial value E down to zero volts.

   Here is the discharge curve of u sub C against time t.

   The differential equation here is R C times du sub C over dt plus u sub C equals zero.

   The solution is u sub C equals E times e to the minus t over tau.

   The same time constant tau, equal to R times C, controls this discharge too.

   After one tau, u sub C drops to about 37 percent of E, during discharge.

   Here is why R and C both matter: they set how fast charge enters or leaves the capacitor.

   A bigger resistor R slows down the current, so charging and discharging both take longer.

   A bigger capacitance C means more charge to move, so it also takes longer.

   So increasing either R or C always makes tau bigger, and the process slower.

   Now let's solve one more worked example, in bac exercise style.

   A capacitor of 50 microfarads discharges through a resistor of 4 kiloohms.

   We first find the time constant tau, using tau equals R times C.

   This calculation gives a time constant of 0{,}2 seconds.

   Now we find u sub C at time t equals 0{,}2 seconds, with E equal to 10 volts.

   Since t equals tau here, this is simply 10 times e to the minus one.

   This calculation gives a voltage of about 3{,}7 volts.

   We frame this result: after one tau, u sub C is about 3{,}7 volts.

   This confirms the 37 percent rule we saw just a moment ago.

   Let's restate the two key results from this whole chapter's final section.

   During discharge, u sub C follows u sub C equals E times e to minus t over tau.

   Increasing R or increasing C always increases tau, and slows down charge or discharge.

   Does increasing the resistance R make the discharge faster or slower?

   This closes our chapter on the capacitor and the RC circuit, from charge to energy to time constants.

   Practice both exercises, convert units carefully, and the exponential laws will become second nature.

   Well done for following this whole chapter through to the end.