ClassX Video Lessons Youtube Channel The Engineering Mindset Capacitor calculations – Basic calculations for capacitors in series and parallel
Capacitors are essential components in many electronic circuits, serving various purposes. This article will guide you through the basic calculations for capacitors in DC circuits, focusing on their behavior in series and parallel configurations.
Capacitors come in different types, with electrolytic and ceramic being the most common. Electrolytic capacitors are polarized, meaning they have a positive and a negative side, which must be connected correctly in a circuit. The negative side is often marked with a dashed line, while the positive side may have a longer lead. However, these leads are usually trimmed during installation, so it’s best not to rely solely on their length for identification. Ceramic capacitors, on the other hand, are non-polarized and can be connected in any direction.
When a capacitor is connected to a DC power supply, the voltage from the battery pushes electrons into the capacitor, charging it to the same voltage as the battery. This charging process is almost instantaneous when connected directly to a battery, but a resistor is often used to delay the charging time. Inside the capacitor, electrons accumulate on one side, creating a voltage difference across the leads. This charge can be held for extended periods and will discharge when a path is provided.
The charge stored in a capacitor is calculated using the formula: Charge (Q) = Capacitance (C) × Voltage (V). For instance, a 12-volt, 100 microfarad capacitor stores a charge of 0.0012 coulombs. To find the required capacitance to store a specific charge, divide the charge by the voltage.
The energy stored in a capacitor is determined by the formula: Energy (E) = 0.5 × Capacitance (C) × Voltage (V)2. Using this formula, a 100 microfarad capacitor charged to 12 volts stores 0.0072 joules of energy.
When capacitors are connected in parallel, their total capacitance is simply the sum of their individual capacitances. For example, if you have a 10 microfarad and a 220 microfarad capacitor in parallel, the total capacitance is 230 microfarads. This configuration is useful for increasing the overall capacitance when a larger capacitor is unavailable.
The total charge in a parallel circuit is calculated as: Total Charge (Q) = Total Capacitance (C) × Voltage (V). For a 9-volt battery and a total capacitance of 230 microfarads, the charge is 0.00207 coulombs.
In a series configuration, the total capacitance is less than the smallest individual capacitor. The formula for total capacitance in series is more complex, but it essentially increases the effective thickness of the insulating material, reducing the overall capacitance. For example, two capacitors of 10 microfarads and 220 microfarads in series result in a total capacitance of 9.56 microfarads.
The charge in a series circuit is the same for each capacitor and is calculated using the same formula as for parallel circuits. However, the voltage across each capacitor will differ, with the smallest capacitor being the limiting factor.
The time constant of a circuit, which determines how quickly a capacitor charges or discharges, is calculated as: Time Constant (τ) = Resistance (R) × Capacitance (C). For example, with a 10 kilo-ohm resistor and a 100 microfarad capacitor, the time constant is 1 second. It takes approximately five time constants for a capacitor to charge to nearly full voltage or discharge to nearly zero.
During charging, the voltage across the capacitor increases exponentially, reaching 63.2% of the battery voltage after one time constant. Similarly, during discharging, the voltage decreases exponentially, reaching 36.8% after one time constant.
Understanding these principles and calculations will help you effectively use capacitors in electronic circuits, whether for energy storage, filtering, or other applications.
Explore more about electronics engineering by checking out additional resources and tutorials. Stay connected with the latest updates by following relevant educational platforms and communities.
Engage in a practical lab session where you will connect capacitors in series and parallel configurations. Measure and record the total capacitance and compare it with theoretical calculations. This hands-on experience will solidify your understanding of how capacitors behave in different configurations.
Utilize simulation software to model circuits with capacitors in series and parallel. Observe the charging and discharging processes, and experiment with different resistor values to see their effect on the time constant. This activity will help you visualize concepts and understand the dynamic behavior of capacitors.
Form small groups and discuss the applications of capacitors in real-world electronic devices. Prepare a short presentation on how capacitors are used in a specific application, such as power supplies or signal filtering. This will enhance your research skills and ability to communicate technical information.
Participate in a workshop where you will solve complex problems involving capacitors in series and parallel. Work through scenarios that require calculating charge, energy, and time constants. This will improve your problem-solving skills and deepen your understanding of capacitor calculations.
Take part in a design challenge where you create a circuit that meets specific criteria using capacitors. Consider factors like total capacitance, charge storage, and discharge time. Present your design and explain your choices, fostering creativity and practical application of theoretical knowledge.
Here’s a sanitized version of the provided YouTube transcript:
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Capacitors are used in many circuits for different purposes, so we’re going to learn some basic capacitor calculations for DC circuits in this video. Capacitors typically look something like this: we have an electrolytic and a ceramic type capacitor. The electrolytic is polarized, meaning one side must be connected to the positive and the other side must be connected to the negative. The ceramic type can generally be connected either way.
On the side of the electrolytic capacitor, we find a dashed line which indicates the negative side. One lead is also longer than the other, indicating the positive side; however, these are normally trimmed down during installation, so you shouldn’t rely on this alone. These two capacitors are represented with symbols. Notice the polarized capacitor has a small plus symbol indicating the positive side.
When connecting to a DC supply, the voltage of the battery will push electrons into the capacitor, and so the capacitor charges up to the same voltage as the battery. Capacitors are charged nearly instantly when connected directly to a battery, but we nearly always use a resistor. This will delay the charging time, and later on in this video, we’ll see how to calculate that.
Inside the capacitor, lots of electrons have built up on one side. They are prevented from moving across due to the insulating material between the two sides. As electrons are negatively charged, we have a buildup of charge on one side compared to the other, resulting in a voltage difference between the two leads. These electrons are held in place, and the capacitor can hold this charge for very long periods of time. When given a path, they will discharge until the capacitor is empty. Electrons do not pass through the capacitor; they simply build up inside and are then released.
The amount of charge stored in a capacitor is calculated using the formula: charge equals capacitance in farads multiplied by the voltage. For example, for a 12-volt, 100 microfarad capacitor, we convert the microfarads to farads and then multiply this by 12 volts to see it stores a charge of 0.0012 coulombs. If we needed to store a charge, we just divide this by the voltage—in this case, 12 volts—to see we need 17 microfarads.
We can calculate the energy stored in a capacitor using the formula: 0.5 multiplied by the capacitance in farads multiplied by the voltage squared. So if this 100 microfarad capacitor was charged to 12 volts, we convert the microfarads to farads and then drop these numbers into the formula to see it is storing 0.0072 joules of energy.
We know that the capacitor will charge up to the voltage of the battery. If we connect a capacitor like this, what will the voltage across the capacitor be? It will be 1.5 volts. If we connect a capacitor like this, what will the voltage be? It will also be 1.5 volts. These are just two different ways to connect capacitors in our circuits: we have series and parallel configurations, which will cause the capacitors to perform differently.
If we place a capacitor in parallel with a lamp, when the battery is removed, the capacitor will begin to power the lamp, which slowly dims as the capacitor discharges. If we use two capacitors, we can power the lamp for longer. Let’s say capacitor 1 is 10 microfarads and capacitor 2 is 220 microfarads. How do we calculate the total capacitance? It’s very simple: the answer is 230 microfarads. The capacitors combine in parallel, so 10 plus 220 equals 230 microfarads.
We can keep adding more, such as a 100 microfarad capacitor, and the total is just the sum of all the capacitors. By placing them in parallel, we are essentially combining these to form a larger capacitor. That’s very useful because if, for example, we needed a large 2000 microfarad capacitor but didn’t have one, we could just use smaller capacitors, such as two 1000 microfarads or four 500 microfarads, etc. It’s also often used for filtering out noise and to provide more current in high-demand circuits.
The total charge stored in parallel circuits is just charge equals the total capacitance multiplied by the voltage. Here we have a 9-volt battery and two capacitors with a total capacitance of 230 microfarads. As this is parallel, one wire is 9 volts and the other is 0 volts, so both capacitors are charged to 9 volts. Therefore, 230 microfarads multiplied by 9 volts will give us 0.00207 coulombs, and with three capacitors, we have 330 microfarads. We multiply this by the 9 volts to get 0.00297 coulombs.
We can also calculate the charge of each capacitor individually using the same formula for each capacitor.
If we place a capacitor in series with a lamp, when we press the switch, it will illuminate but then become dimmer as the capacitor reaches the voltage level of the battery. Once it achieves this, the lamp will be off. Remember, electrons do not flow through a capacitor because of the insulating material inside. The electrons are simply accumulating inside one of the plates, and as they accumulate, they are rejecting an equal amount of electrons off of the opposite plate. So a current can only flow when the capacitor charges or discharges.
Currently, with the battery removed, there is no way for the capacitor to discharge, so it will hold the voltage at the same level. It doesn’t matter if we connect or disconnect the battery; the lamp will not turn on. However, if we provide another path, when the switch is pressed, the capacitor can now discharge, allowing the electrons to flow through the lamp and illuminate it. This will become dimmer as the capacitor discharges.
What if we had two capacitors connected in series? Again, capacitor 1 is 10 microfarads and capacitor 2 is 220 microfarads. How do we find the total capacitance? We use this formula. It might look difficult, but it’s actually very simple. All we need to do is input our capacitor values of 10 and 220 microfarads. We can type it like this on our calculator or into Excel, which gives a total of 9.56 microfarads. Notice that the total capacitance is now smaller than the lowest value capacitor. If we added a third capacitor of 100 microfarads to the circuit, we get a total capacitance of 8.73 microfarads, so it has decreased even more. That’s because by combining these in series, we’re essentially increasing the thickness of the insulating material, weakening the attraction of the negatively charged electrons to the positively charged holes on the opposite side.
The total charge of the series capacitors is found using the formula: charge equals capacitance in farads multiplied by the voltage. If we use a 9-volt battery, we convert the microfarads to farads and see the total charge equals 0.008604 coulombs. The total charge for the three series capacitor circuit is therefore 0.00007857 coulombs. The charge held by each capacitor individually is very easy to calculate in series circuits because it’s the same as the total charge. Each capacitor holds the same number of electrons when in series, as the current was exactly the same in all parts of the circuit. The same number of electrons that were pushed into one plate were pushed out of the opposite plate, so each series capacitor can only ever be charged to the same level. The smallest capacitor will therefore be the limiting factor. However, because each capacitor can hold a different capacity, the voltage of each capacitor will be different.
We find the voltage of each capacitor using the formula: voltage equals charge in coulombs divided by the capacity in farads. For this circuit, we see that capacitor 1 is 7.8 volts, capacitor 2 is 0.35 volts, and capacitor 3 is 0.78 volts. These combine to the total voltage of the battery, which is 9 volts.
Let’s say we have a 9-volt battery, a 100 microfarad capacitor, a 10 kilo-ohm resistor, and a switch, which are all in series. The capacitor is fully discharged, and we read zero volts across the two leads. When we close the switch, the capacitor will charge. The voltage will increase until it is the same level as the battery. The voltage increase is not instant; it will have an exponential curve. At first, the voltage increases rapidly and then slows down until it reaches the same voltage level as the battery.
We split this curve into six segments, but we’re only interested in the first five because at the fifth marker, we’re basically at full voltage, so we can ignore anything past this. Each segment represents something called a time constant. Therefore, as we have five segments, we have five time constants, so it will take five time constants to charge the capacitor from zero to just under one hundred percent.
To calculate the time constant, we use this formula: time constant in seconds equals the resistance in ohms multiplied by the capacitance in farads. We convert our resistor to ohms and our capacitor value to farads, and we see that 10,000 ohms multiplied by 0.0001 farads equals one. So in this example, the time constant is equal to one second. Therefore, five of these is five seconds, meaning it takes five seconds for the capacitor to fully charge to 9 volts. If the resistor was just 1000 ohms, the time constant would be 0.1 seconds, so it would take 0.5 seconds to reach 9 volts. If the capacitor was 1000 microfarads, it would take 50 seconds in total.
Coming back to our original circuit, we can therefore calculate the voltage level at each time constant. At point 1, the voltage is always 63.2 percent; at point 2, it is 86.5; at point 3, it is 95; at point 4, it is 98.2 percent; and at point 5, it is 99.3 percent. So the voltage will never actually reach 100 percent; that’s also why we stop at just five points.
In this example, after one second, the capacitor voltage is 5.68 volts; after two seconds, it’s 7.78 volts; after three seconds, it’s 8.55 volts; after four seconds, it’s 8.83 volts; and after five seconds, it’s 8.94 volts. If you needed a more precise answer, we could also calculate each point like this.
Remember, because this is in series, the current of the circuit decreases while the voltage of the capacitor increases. Once at full voltage, no current will flow in the circuit. If the resistor was a lamp, it would therefore instantly reach full brightness when the switch was closed but then become dimmer as the capacitor reaches full voltage. When we provide a path for the capacitor to discharge, the electrons will leave the capacitor, and the voltage of the capacitor reduces. It doesn’t discharge instantly but follows an exponential curve.
We split this curve into six segments, but again we’re only interested in the first five. At point 1, the voltage is always 36.8 percent; point 2 will be 13.5 percent; point 3 will be 5 percent; point 4 will be 1.8 percent; and point 5 will be 0.7 percent.
For example, if we had a 9-volt battery, a lamp with a resistance of 500 ohms, and a 2000 microfarad capacitor, our time constant would be 500 ohms multiplied by 0.002 farads, which is one second. So at the very moment the battery is disconnected, the capacitor will be at 9 volts, and as it powers the circuit, the lamp will also experience 9 volts. After one time constant, in this case 1 second, the voltage will be 36.8 percent, which is 3.312 volts. At 2 seconds, it’s 1.215 volts; at 3 seconds, it’s 0.45 volts; at 4 seconds, it’s 0.162 volts; and at 5 seconds, it’s 0.063 volts. So the lamp will be illuminated for just under 3 seconds, obviously becoming dimmer towards the end of the three seconds.
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This version removes any informal language and ensures clarity while maintaining the technical content.
Capacitor – A device used in electrical circuits to store energy in an electric field, typically consisting of two conductive plates separated by an insulating material. – The capacitor in the circuit was used to smooth out voltage fluctuations.
Capacitance – The ability of a system to store an electric charge, measured in farads. – The capacitance of the capacitor was increased by using a dielectric material with a higher permittivity.
Voltage – The electric potential difference between two points, which drives the flow of electric current in a circuit. – The voltage across the resistor was measured to determine the current flowing through the circuit.
Charge – A property of matter that causes it to experience a force when placed in an electromagnetic field, measured in coulombs. – The charge on the capacitor plates was calculated using the formula Q = CV.
Energy – The capacity to do work, which in electrical systems is often stored or transferred by electric fields or currents. – The energy stored in the capacitor was released when the circuit was closed.
Series – A configuration of circuit components connected end-to-end, so that the same current flows through each component. – The resistors were connected in series, resulting in a higher total resistance.
Parallel – A configuration of circuit components connected across the same voltage source, allowing current to divide among the paths. – The capacitors were arranged in parallel to increase the total capacitance of the circuit.
Resistance – A measure of the opposition to the flow of electric current, resulting in the conversion of electrical energy into heat, measured in ohms. – The resistance of the wire was calculated to ensure it could safely carry the current.
Time – A continuous, measurable quantity in which events occur in a sequence, often used in physics to describe the duration of processes. – The time constant of the RC circuit was determined by the product of resistance and capacitance.
Electrons – Subatomic particles with a negative charge, which flow through conductors to create electric current. – The movement of electrons in the conductor was analyzed to understand the current flow in the circuit.
