ClassX Video Lessons Youtube Channel The Engineering Mindset Capacitor discharge time – how to calculate with examples
When a capacitor discharges, it releases stored electrons, causing its voltage to decrease. This process doesn’t happen instantly; instead, it follows an exponential decay curve. For simplicity, we divide this curve into six segments, focusing primarily on the first five.
At the first segment, the voltage drops to 36.8% of its initial value. By the second segment, it falls to 13.5%. At the third, it reaches 5%, then 1.8% at the fourth, and finally 0.7% at the fifth.
Consider a circuit with a 9-volt battery, a lamp with a resistance of 500 ohms, and a 2000 microfarad capacitor. The time constant, a key factor in this process, is calculated by multiplying the resistance (500 ohms) by the capacitance (0.002 farads), resulting in one second.
Initially, when the battery is disconnected, the capacitor holds 9 volts. As it discharges, the lamp experiences the same voltage. After one time constant (one second), the voltage drops to 36.8%, or 3.312 volts. At two seconds, it’s 1.215 volts; at three seconds, 0.45 volts; at four seconds, 0.162 volts; and at five seconds, 0.063 volts. The lamp will shine for nearly three seconds, dimming as time progresses.
Now, let’s examine a different scenario: a 9-volt battery, a 100 microfarad capacitor, a 10 kilo-ohm resistor, and a switch, all connected in series. Initially, the capacitor is fully discharged, showing zero volts across its leads. Upon closing the switch, the capacitor begins to charge, with its voltage rising until it matches the battery’s voltage.
This charging process also follows an exponential curve, initially increasing rapidly before slowing down. We again divide this curve into six segments, focusing on the first five, as the capacitor is nearly fully charged by the fifth segment.
The time constant is determined by multiplying the resistance (in ohms) by the capacitance (in farads). For our example, 10,000 ohms multiplied by 0.0001 farads equals one second. Thus, it takes five seconds for the capacitor to charge to nearly 9 volts. If the resistance were only 1000 ohms, the time constant would be 0.1 seconds, requiring just 0.5 seconds to reach 9 volts. With a 1000 microfarad capacitor, the total time would extend to 50 seconds.
As the capacitor’s size increases, so does the time required for charging. Similarly, higher resistance values lead to longer charging times.
Returning to our original circuit, we can calculate the voltage at each time constant. At the first point, the voltage is 63.2%; at the second, 86.5%; at the third, 95%; at the fourth, 98.2%; and at the fifth, 99.3%. The voltage never quite reaches 100%, which is why we stop at five points.
In this example, after one second, the capacitor voltage is 5.68 volts; after two seconds, 7.78 volts; after three seconds, 8.55 volts; after four seconds, 8.83 volts; and after five seconds, 8.94 volts.
For more precise calculations, you can determine each point individually. In a series circuit, as the capacitor’s voltage increases, the circuit’s current decreases. Once the capacitor is fully charged, no current flows through the circuit.
If the resistor were a lamp, it would initially shine brightly when the switch is closed, then gradually dim as the capacitor reaches full voltage.
For further exploration of electronics engineering, consider watching related videos or visiting educational websites like engineeringmindset.com.
Engage with an online simulation tool to visualize the exponential decay of voltage in a discharging capacitor. Adjust parameters such as resistance and capacitance to see their effects on the discharge curve. This hands-on activity will help you understand the relationship between these variables and the discharge time.
Form small groups to discuss the concept of time constants in capacitor discharge. Each group will prepare a short presentation explaining how the time constant affects the discharge process, using real-world examples. This collaborative activity will enhance your understanding through peer learning and communication.
Conduct a lab experiment where you discharge a capacitor through a resistor and measure the voltage at different time intervals. Plot the results to observe the exponential decay curve. This practical experience will reinforce theoretical concepts and improve your experimental skills.
Participate in a workshop where you solve various problems related to capacitor discharge and charging. Work through calculations of time constants, voltage levels, and current changes. This activity will strengthen your analytical skills and deepen your understanding of the mathematical aspects of capacitors.
Engage in a role-play activity where you simulate the process of charging and discharging a capacitor. Assign roles such as electrons, resistors, and capacitors to understand the flow of current and voltage changes. This creative approach will help you visualize and remember the processes involved.
Here’s a sanitized version of the provided YouTube transcript:
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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 we’re only interested in the first five.
At point one, the voltage is always 36.8%. At point two, it will be 13.5%. At point three, it will be 5%. At point four, it will be 1.8%, and at point five, it will be 0.7%.
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 equals one second.
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, one second), the voltage will be 36.8%, 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 three seconds, becoming dimmer towards the end of that time.
Now, let’s say we have a 9-volt battery, a 100 microfarad capacitor, a 10 kilo-ohm resistor, and a switch, 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, and the voltage will increase until it matches the battery voltage.
The voltage increase is not instant; it will follow an exponential curve. Initially, 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, focusing on the first five, as at the fifth marker, we’re essentially at full voltage.
Each segment represents a time constant. Therefore, with five segments, we have five time constants. It will take five time constants to charge the capacitor from zero to just under 100%.
To calculate the time constant, we use the formula: time constant in seconds equals resistance in ohms multiplied by capacitance in farads. Converting our resistor to ohms and our capacitor value to farads, we see that 10,000 ohms multiplied by 0.0001 farads equals 1 second.
Thus, the time constant is one second, meaning it takes five seconds for the capacitor to fully charge to 9 volts. If the resistor were 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 were 1000 microfarads, it would take 50 seconds in total.
As the capacitor size increases, the time taken will also increase. If the resistor value increases, then the time taken also increases.
Returning to our original circuit, we can calculate the voltage level at each time constant. At point 1, the voltage is always 63.2%. At point 2, it is 86.5%. At point 3, it is 95%. At point 4, it is 98.2%, and at point 5, it is 99.3%. The voltage will never actually reach 100%, which is 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 8.4 volts.
If you need 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 were a lamp, it would instantly reach full brightness when the switch was closed but then become dimmer as the capacitor reaches full voltage.
Check out one of the videos on screen now to continue learning about electronics engineering. This is the end of this video. Don’t forget to follow us on social media and visit the engineeringmindset.com.
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This version removes any informal language and maintains a professional tone while preserving 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.
Voltage – The electric potential difference between two points, which causes current to flow in a circuit. – The voltage across the resistor was measured to determine the current flowing through the circuit.
Resistance – A measure of the opposition to the flow of electric current in a conductor, typically measured in ohms. – The resistance of the wire was calculated to ensure it could safely carry the required 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 taken for the capacitor to fully charge was recorded during the experiment.
Constant – A value that does not change in a given context, often used to describe fixed parameters in equations or systems. – The dielectric constant of the material was crucial in determining the capacitance of the capacitor.
Charging – The process of storing energy in a capacitor by applying a voltage across its terminals, causing an accumulation of electric charge. – During the charging phase, the voltage across the capacitor increased exponentially.
Discharging – The process of releasing stored energy from a capacitor, resulting in a decrease of voltage across its terminals. – The discharging of the capacitor was monitored to study the decay of voltage over time.
Circuit – A closed loop through which electric current flows, consisting of various electrical components such as resistors, capacitors, and power sources. – The circuit was designed to test the efficiency of different capacitor materials.
Electrons – Subatomic particles with a negative charge that flow through conductors to create electric current. – The movement of electrons in the conductor was analyzed to understand the current flow in the circuit.
Farads – The unit of capacitance in the International System of Units, defined as one coulomb of charge per volt. – The capacitance of the capacitor was measured in farads to determine its ability to store charge.
