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Experiment:
Charging and Discharging the 100 µF Capacitor with a 330 Ω Resistor
Circuit Diagram
Materials Required i. ii. iii. iv. v. vi.
Breadboard -‐ 1 Capacitor: 100 µF -‐ 1 LED -‐ 1 Resistor: 330 Ω-‐1 Colour Code: 330 Ω -‐ Orange Orange Brown Gold 9 V BaHery -‐ 1 ConnecJng Wire Pieces
Points to • In this experiment, we need to observe how the voltage across the capacitor changes as it gets charged. For conJnuous voltage measurement, we set the meter probe across the capacitor terminals, before we give supply to the circuit. • Before starJng the experiment ensure that the capacitor is fully discharged. We can discharge the capacitor instantly by shorJng its two legs.
Step No. 1
Take a breadboard.
Step No. 2
Connect a 330 Ω resistor on the breadboard.
Step No. 3
Connect the posiJve terminal of a 100 µF capacitor to the right leg of the resistor and the negaJve terminal to any different column of the breadboard.
Step No. 4
Connect the posiJve terminal of an LED to the negaJve terminal of the capacitor and the negaJve terminal to any different column of the breadboard.
Step No. 5
Connect the leT leg of the resistor to the first row of the breadboard and the negaJve terminal of the LED to the second row of the breadboard.
Step No. 6
Connect two wires to the mulJmeter probes as shown in figure below.
Step No. 7
Insert the wires aHached to the meter probes in parallel to the capacitor in such a way that the red and black wires of the meter are connected to the posiJve and negaJve terminals of the capacitor, respecJvely.
Step No. 8
Connect the negaJve terminal of the baHery to the second row of the breadboard.
Step No. 9
Connect the posiJve terminal of the baHery to the first row of the breadboard. As we complete the circuit, the LED glows brightly and the meter displays a voltage reading approximately equal to zero volt.
ObservaLon Step No. 8 shows the instantaneous voltage reading of 0.06 V across the capacitor at t = 0+ second (immediately aTer the connecJon is complete). The LED glows with maximum intensity.
Reasoning At t = 0 (iniJal state), the capacitor has no charge. An uncharged capacitor behaves like a short circuit. So, the current at this point of Jme is maximum and depends upon the net resistance offered by the LED and the resistor.
Step No. 10
The meter will show a conJnuously increasing voltage reading. The intensity of the LED glow will decrease with Jme.
ObservaLon Step No. 9 shows the instantaneous voltage reading of 5.71 V across the capacitor between t = 0 and t = 1 second. The voltage reading is seen to be rising on the meter. The intensity with which the LED glows decreases gradually.
Reasoning The intensity with which the LED glows decreases because the current in the circuit decreases. In this case, the charged capacitor opposes the baHery voltage. Hence, the net voltage in the circuit is the difference between the baHery and capacitor voltages. As the voltage across the capacitor increases, the net voltage in the circuit decreases. At any instant, Net Voltage = BaHery Voltage − Voltage across the capacitor = 9 − 5.48 = 3.52 V As the capacitor gets charged, the net voltage will decrease further and so will the current.
Step No. 11
The voltage reading becomes steady aTer someJme. The LED is OFF.
ObservaLon Step No. 11 shows the instantaneous voltage reading of 8.03 V aTer some Jme has elapsed. The LED of OFF, and the meter reading does not increase any more. This state is called the steady state of the capacitor. In steady state, the capacitor aHains a voltage which is very close to the baHery voltage. Thus, it gets almost fully charged.
Reasoning From the observaJon, it is clear that the voltage across the capacitor is 8.03 V. Net Voltage = BaHery Voltage − Voltage across the capacitor = 9 − 8.03 = 0.97 V We know that an LED requires a minimum voltage of 1.8-‐2 V across its terminals to allow current to through it. Since the net voltage is less than the minimum voltage required by the LED, it turns OFF. The net current in the circuit is zero.
ObservaLon • The rate of increment of the voltage reading is iniJally fast but subsequently slows down with Jme. • The LED glows and then slowly fades out. • The meter will show a conJnuously increasing voltage reading. The reading becomes steady aTer someJme.
Result • The circuit has the maximum amount of current when the capacitor is uncharged. As the capacitor starts gaining charge, the current in the circuit starts decreasing. • When the capacitor is fully charged, it offers infinite resistance to the direct current. The current in the circuit becomes zero. This is called the steady state of the capacitor. A capacitor will never get fully charged, or we can say that it will be charged in infinite Jme. • The capacitor does not charge at a constant rate. The charging is iniJally fast, but starts slowing down with Jme.
For Jme constant calculaJons during charging: Refer the Manual Experiment: Charging the Capacitor with a Resistor
Next Step •
•
Remove the charged capacitor from the circuit and keep it aside. We will now discharge the charged capacitor in the discharging experiment. Remove all the circuit components and jump to step no. 12
Step No. 12
Connect a 330 Ω resistor and an LED in series on the breadboard.
Step No. 13
Connect red and black wires of the meter to the leT leg of the resistor and the negaJve terminal of the LED, respecJvely.
ObservaLon Step No. 13 shows the instantaneous voltage reading of 0.00 V across the capacitor at t = 0+ second (immediately aTer the connecJon is complete).
Reasoning Even before the meter measures the voltage reading, the capacitor starts discharging at t = 0+ second. The LED glows because the capacitor acts as a temporary baHery and discharges through the LED and the resistor, producing a current in the circuit. The moment the capacitor starts discharging, the current in the circuit is maximum. Over Jme, the current decreases in the circuit and eventually becomes zero.
Reasoning There may be an instant when the LED does not glow, and one can wrongly assume that the current in the circuit is zero. This may happen due to low current in the circuit. Thus, it is important to know the amount of current flowing in the circuit at any given instant to avoid misinterpretaJons.
CalculaLon CalculaLng current at t = 0+ second: From Step No. 11, we know that the voltage across the charged capacitor = 8.03 V-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐(1) From Ohm’s law, we can calculate the iniJal current (at t = 0+ second). At any instant, Voltage across capacitor = Voltage across LED + Voltage across resistor-‐-‐-‐-‐-‐-‐-‐-‐(2) A red LED needs a minimum voltage of 1.8-‐2.1 V to glow. When it glows brightly, a voltage drop of 2 V occurs its terminals.
CalculaLon Voltage drop across LED (red) = 2 V -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐(3) From (1), (2) and (3), Voltage across resistor = 8.03 − 2 = 6.03 V Resistance of resistor = 330 Ω From Ohm’s law: Current = Voltage/Resistance = 6.03/330 = 18.27 mA Hence, iniJal current in the circuit = 18.27 mA.
Step No. 14
Connect the charged capacitor in parallel to the series combinaJon of the LED and the resistor in such a way that its posiJve and negaJve terminals are connected to the leT leg of the resistor and the negaJve terminal of the LED, respecJvely. The LED glows for a moment. The meter displays a voltage reading approximately equal to baHery voltage.
Step No. 15
We can observe on the display that the voltage reading falls rapidly and the LED slowly fades out.
ObservaLon Step No. 15 shows the instantaneous voltage reading of 2.39 V across the capacitor between t = 0 and t = 1 second. The LED is almost OFF.
Reasoning This may happen due to two reasons: i) The current in the circuit is very low. The current determines the brightness of the LED. More current means more light. ii) The current in the circuit is zero. So, let us calculate the current in the circuit to check whether it is low or zero.
CalculaLon We know that: Voltage across capacitor = Voltage drop across LED + Voltage across resistor Minimum voltage drop across LED (red) = 2 V From Step No. 11, voltage across capacitor = 2.39 V Voltage across resistor = 2.39 − 2 = 0.39 V From Ohm’s law, Current = Voltage/Resistance = 0.39/330 = 1.18 mA The current in the circuit is very low, and hence the intensity with which the LED glows is low.
Step No. 16
Figure shows the instantaneous voltage reading across the capacitor when it is almost fully discharged. We must note that the voltage across the capacitor is close to zero, and not zero. A capacitor will never get fully discharged. The LED is OFF.
ObservaLon Step No. 16 shows the instantaneous voltage reading of 0.58 V across the capacitor when it is almost fully discharged. We must note that the voltage across the capacitor is close to zero, and not zero. A capacitor will never get fully discharged. The LED is OFF.
Reasoning The LED is OFF because the current in the circuit is zero.
ObservaLon •
The meter will show a conJnuously decreasing voltage. The reading becomes steady aTer someJme.
•
The rate of decrease of voltage is iniJally fast but subsequently slows down with Jme.
•
The LED glows and then slowly fades out.
Result • The moment the capacitor starts discharging, the current in the circuit is maximum. As the capacitor discharges and loses voltage, the current in the circuit decreases. • When the capacitor is fully discharged, it behaves like an open circuit. The current in the circuit becomes zero. • A capacitor will fully discharged in infinite Jme. • The discharging is iniJally fast, and subsequently slows down with Jme.
For Jme constant calculaJons during discharging: Refer the Manual Experiment: Discharging the Capacitor with a Resistor
Project Report To write project report, refer the following experiments from the manual: a) Charging the Capacitor with a Resistor b) Discharging the Capacitor with a Resistor In the report, do include the following: a) Circuit diagram, Circuit ExplanaJon b) Final circuit picture c) Pictures of ‘MulJmeter readings’: Step No. 8, Step No. 9, Step No. 10 and Step No. 11 d) ObservaJon e) Reasoning (if any) f) Result (if any) g) General Theory: About components used (LED, resistor, mulJmeter, DC baHery, capacitor).
TroubleshooLng Tips • •
• • •
Ensure that the baHery voltage is more than 6 volt. Ensure that the wires of the baHery connector are properly inserted into the breadboard. The red wire should be inserted into the first row, and the black wire into the second row of the breadboard. Ensure that the LED is in working state using a mulJmeter. Ensure that the capacitor is fully discharged before the start of the experiment. Ensure that there are no loose connecJons.
Result PrecauLon Notes
Circuit Measure AcLvity Advantages ModificaLon
Do you know?
Charging and Discharging the 100 µF Capacitor with a 330 Ω Resistor
Circuit Diagram
Materials Required i. ii. iii. iv. v. vi.
Breadboard -‐ 1 Capacitor: 100 µF -‐ 1 LED -‐ 1 Resistor: 330 Ω-‐1 Colour Code: 330 Ω -‐ Orange Orange Brown Gold 9 V BaHery -‐ 1 ConnecJng Wire Pieces
Points to • In this experiment, we need to observe how the voltage across the capacitor changes as it gets charged. For conJnuous voltage measurement, we set the meter probe across the capacitor terminals, before we give supply to the circuit. • Before starJng the experiment ensure that the capacitor is fully discharged. We can discharge the capacitor instantly by shorJng its two legs.
Step No. 1
Take a breadboard.
Step No. 2
Connect a 330 Ω resistor on the breadboard.
Step No. 3
Connect the posiJve terminal of a 100 µF capacitor to the right leg of the resistor and the negaJve terminal to any different column of the breadboard.
Step No. 4
Connect the posiJve terminal of an LED to the negaJve terminal of the capacitor and the negaJve terminal to any different column of the breadboard.
Step No. 5
Connect the leT leg of the resistor to the first row of the breadboard and the negaJve terminal of the LED to the second row of the breadboard.
Step No. 6
Connect two wires to the mulJmeter probes as shown in figure below.
Step No. 7
Insert the wires aHached to the meter probes in parallel to the capacitor in such a way that the red and black wires of the meter are connected to the posiJve and negaJve terminals of the capacitor, respecJvely.
Step No. 8
Connect the negaJve terminal of the baHery to the second row of the breadboard.
Step No. 9
Connect the posiJve terminal of the baHery to the first row of the breadboard. As we complete the circuit, the LED glows brightly and the meter displays a voltage reading approximately equal to zero volt.
ObservaLon Step No. 8 shows the instantaneous voltage reading of 0.06 V across the capacitor at t = 0+ second (immediately aTer the connecJon is complete). The LED glows with maximum intensity.
Reasoning At t = 0 (iniJal state), the capacitor has no charge. An uncharged capacitor behaves like a short circuit. So, the current at this point of Jme is maximum and depends upon the net resistance offered by the LED and the resistor.
Step No. 10
The meter will show a conJnuously increasing voltage reading. The intensity of the LED glow will decrease with Jme.
ObservaLon Step No. 9 shows the instantaneous voltage reading of 5.71 V across the capacitor between t = 0 and t = 1 second. The voltage reading is seen to be rising on the meter. The intensity with which the LED glows decreases gradually.
Reasoning The intensity with which the LED glows decreases because the current in the circuit decreases. In this case, the charged capacitor opposes the baHery voltage. Hence, the net voltage in the circuit is the difference between the baHery and capacitor voltages. As the voltage across the capacitor increases, the net voltage in the circuit decreases. At any instant, Net Voltage = BaHery Voltage − Voltage across the capacitor = 9 − 5.48 = 3.52 V As the capacitor gets charged, the net voltage will decrease further and so will the current.
Step No. 11
The voltage reading becomes steady aTer someJme. The LED is OFF.
ObservaLon Step No. 11 shows the instantaneous voltage reading of 8.03 V aTer some Jme has elapsed. The LED of OFF, and the meter reading does not increase any more. This state is called the steady state of the capacitor. In steady state, the capacitor aHains a voltage which is very close to the baHery voltage. Thus, it gets almost fully charged.
Reasoning From the observaJon, it is clear that the voltage across the capacitor is 8.03 V. Net Voltage = BaHery Voltage − Voltage across the capacitor = 9 − 8.03 = 0.97 V We know that an LED requires a minimum voltage of 1.8-‐2 V across its terminals to allow current to through it. Since the net voltage is less than the minimum voltage required by the LED, it turns OFF. The net current in the circuit is zero.
ObservaLon • The rate of increment of the voltage reading is iniJally fast but subsequently slows down with Jme. • The LED glows and then slowly fades out. • The meter will show a conJnuously increasing voltage reading. The reading becomes steady aTer someJme.
Result • The circuit has the maximum amount of current when the capacitor is uncharged. As the capacitor starts gaining charge, the current in the circuit starts decreasing. • When the capacitor is fully charged, it offers infinite resistance to the direct current. The current in the circuit becomes zero. This is called the steady state of the capacitor. A capacitor will never get fully charged, or we can say that it will be charged in infinite Jme. • The capacitor does not charge at a constant rate. The charging is iniJally fast, but starts slowing down with Jme.
For Jme constant calculaJons during charging: Refer the Manual Experiment: Charging the Capacitor with a Resistor
Next Step •
•
Remove the charged capacitor from the circuit and keep it aside. We will now discharge the charged capacitor in the discharging experiment. Remove all the circuit components and jump to step no. 12
Step No. 12
Connect a 330 Ω resistor and an LED in series on the breadboard.
Step No. 13
Connect red and black wires of the meter to the leT leg of the resistor and the negaJve terminal of the LED, respecJvely.
ObservaLon Step No. 13 shows the instantaneous voltage reading of 0.00 V across the capacitor at t = 0+ second (immediately aTer the connecJon is complete).
Reasoning Even before the meter measures the voltage reading, the capacitor starts discharging at t = 0+ second. The LED glows because the capacitor acts as a temporary baHery and discharges through the LED and the resistor, producing a current in the circuit. The moment the capacitor starts discharging, the current in the circuit is maximum. Over Jme, the current decreases in the circuit and eventually becomes zero.
Reasoning There may be an instant when the LED does not glow, and one can wrongly assume that the current in the circuit is zero. This may happen due to low current in the circuit. Thus, it is important to know the amount of current flowing in the circuit at any given instant to avoid misinterpretaJons.
CalculaLon CalculaLng current at t = 0+ second: From Step No. 11, we know that the voltage across the charged capacitor = 8.03 V-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐(1) From Ohm’s law, we can calculate the iniJal current (at t = 0+ second). At any instant, Voltage across capacitor = Voltage across LED + Voltage across resistor-‐-‐-‐-‐-‐-‐-‐-‐(2) A red LED needs a minimum voltage of 1.8-‐2.1 V to glow. When it glows brightly, a voltage drop of 2 V occurs its terminals.
CalculaLon Voltage drop across LED (red) = 2 V -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐(3) From (1), (2) and (3), Voltage across resistor = 8.03 − 2 = 6.03 V Resistance of resistor = 330 Ω From Ohm’s law: Current = Voltage/Resistance = 6.03/330 = 18.27 mA Hence, iniJal current in the circuit = 18.27 mA.
Step No. 14
Connect the charged capacitor in parallel to the series combinaJon of the LED and the resistor in such a way that its posiJve and negaJve terminals are connected to the leT leg of the resistor and the negaJve terminal of the LED, respecJvely. The LED glows for a moment. The meter displays a voltage reading approximately equal to baHery voltage.
Step No. 15
We can observe on the display that the voltage reading falls rapidly and the LED slowly fades out.
ObservaLon Step No. 15 shows the instantaneous voltage reading of 2.39 V across the capacitor between t = 0 and t = 1 second. The LED is almost OFF.
Reasoning This may happen due to two reasons: i) The current in the circuit is very low. The current determines the brightness of the LED. More current means more light. ii) The current in the circuit is zero. So, let us calculate the current in the circuit to check whether it is low or zero.
CalculaLon We know that: Voltage across capacitor = Voltage drop across LED + Voltage across resistor Minimum voltage drop across LED (red) = 2 V From Step No. 11, voltage across capacitor = 2.39 V Voltage across resistor = 2.39 − 2 = 0.39 V From Ohm’s law, Current = Voltage/Resistance = 0.39/330 = 1.18 mA The current in the circuit is very low, and hence the intensity with which the LED glows is low.
Step No. 16
Figure shows the instantaneous voltage reading across the capacitor when it is almost fully discharged. We must note that the voltage across the capacitor is close to zero, and not zero. A capacitor will never get fully discharged. The LED is OFF.
ObservaLon Step No. 16 shows the instantaneous voltage reading of 0.58 V across the capacitor when it is almost fully discharged. We must note that the voltage across the capacitor is close to zero, and not zero. A capacitor will never get fully discharged. The LED is OFF.
Reasoning The LED is OFF because the current in the circuit is zero.
ObservaLon •
The meter will show a conJnuously decreasing voltage. The reading becomes steady aTer someJme.
•
The rate of decrease of voltage is iniJally fast but subsequently slows down with Jme.
•
The LED glows and then slowly fades out.
Result • The moment the capacitor starts discharging, the current in the circuit is maximum. As the capacitor discharges and loses voltage, the current in the circuit decreases. • When the capacitor is fully discharged, it behaves like an open circuit. The current in the circuit becomes zero. • A capacitor will fully discharged in infinite Jme. • The discharging is iniJally fast, and subsequently slows down with Jme.
For Jme constant calculaJons during discharging: Refer the Manual Experiment: Discharging the Capacitor with a Resistor
Project Report To write project report, refer the following experiments from the manual: a) Charging the Capacitor with a Resistor b) Discharging the Capacitor with a Resistor In the report, do include the following: a) Circuit diagram, Circuit ExplanaJon b) Final circuit picture c) Pictures of ‘MulJmeter readings’: Step No. 8, Step No. 9, Step No. 10 and Step No. 11 d) ObservaJon e) Reasoning (if any) f) Result (if any) g) General Theory: About components used (LED, resistor, mulJmeter, DC baHery, capacitor).
TroubleshooLng Tips • •
• • •
Ensure that the baHery voltage is more than 6 volt. Ensure that the wires of the baHery connector are properly inserted into the breadboard. The red wire should be inserted into the first row, and the black wire into the second row of the breadboard. Ensure that the LED is in working state using a mulJmeter. Ensure that the capacitor is fully discharged before the start of the experiment. Ensure that there are no loose connecJons.
Result PrecauLon Notes
Circuit Measure AcLvity Advantages ModificaLon
Do you know?
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