Experiment:Charging and Discharging the 100 F Capacitor with a 330 Resistor

Circuit Diagram

i. Breadboard - 1ii. Capacitor: 100 F - 1iii. LED - 1iv. Resistor: 330 -1 Colour Code: 330 - Orange Orange Brown Goldv. 9 V BaHery - 1vi. ConnecJng Wire Pieces

Materials Required

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.

Points to Remember

Take a breadboard.

Step No. 1

Connect a 330 resistor on the breadboard.

Step No. 2

Connect the posiJve terminal of a 100 F capacitor to the right leg of the resistor and the negaJve terminal to any dierent column of the breadboard.

Step No. 3

Connect the posiJve terminal of an LED to the negaJve terminal of the capacitor and the negaJve terminal to any dierent column of the breadboard.

Step No. 4

Connect the leT leg of the resistor to the rst row of the breadboard and the negaJve terminal of the LED to the second row of the breadboard.

Step No. 5

Connect two wires to the mulJmeter probes as shown in gure below.

Step No. 6

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. 7

Connect the negaJve terminal of the baHery to the second row of the breadboard.

Step No. 8

Connect the posiJve terminal of the baHery to the rst 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.

Step No. 9

ObservaLonStep 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.

ReasoningAt 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 oered by the LED and the resistor.

The meter will show a conJnuously increasing voltage reading. The intensity of the LED glow will decrease with Jme.

Step No. 10

ObservaLonStep 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.

ReasoningThe 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 dierence 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 VAs the capacitor gets charged, the net voltage will decrease further and so will the current.

The voltage reading becomes steady aTer someJme. The LED is OFF.

Step No. 11

ObservaLonStep 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.

ReasoningFrom 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 pass 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.

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.

ObservaLon

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 oers innite 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 innite Jme.

The capacitor does not charge at a constant rate. The charging is iniJally fast, but starts slowing down with Jme.

Result

For Jme constant calculaJons during

charging: Refer the ManualExperiment: Charging the Capacitor with a Resistor

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

Next Step

Connect a 330 resistor and an LED in series on the breadboard.

Step No. 12

Connect red and black wires of the meter to the leT leg of the resistor and the negaJve terminal of the LED, respecJvely.

Step No. 13

ObservaLonStep No. 13 shows the instantaneous voltage reading of 0.00 V across the capacitor at t = 0+ second (immediately aTer the connecJon is complete).

ReasoningEven 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.

ReasoningThere 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 owing in the circuit at any given instant to avoid misinterpretaJons.

CalculaLonCalculaLng current at t = 0+ second:From Step No. 11, we know that the voltage across the charged capacitor = 8.03 V----------(1)

From Ohms 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.

CalculaLonVoltage drop across LED (red) = 2 V -------------(3)

From (1), (2) and (3),Voltage across resistor = 8.03 2 = 6.03 VResistance of resistor = 330

From Ohms law:Current = Voltage/Resistance = 6.03/330 = 18.27 mAHence, iniJal current in the circuit = 18.27 mA.

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. 14

We can observe on the display that the voltage reading falls rapidly and the LED slowly fades out.

Step No. 15

ObservaLonStep 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.

ReasoningThis 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.

CalculaLonWe know that:Voltage across capacitor = Voltage drop across LED + Voltage across resistorMinimum voltage drop across LED (red) = 2 VFrom Step No. 11, voltage across capacitor = 2.39 VVoltage across resistor = 2.39 2 = 0.39 V

From Ohms law, Current = Voltage/Resistance = 0.39/330

= 1.18 mAThe current in the circuit is very low, and hence the intensity with which the LED glows is low.

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.

Step No. 16

ObservaLonStep 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.

ReasoningThe LED is OFF because the current in the circuit is zero.

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.

ObservaLon

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 innite Jme. The discharging is iniJally fast, and subsequently slows

down with Jme.

Result

For Jme constant calculaJons during

discharging: Refer the ManualExperiment: Discharging the Capacitor with a Resistor

To write project report, refer the following experiments from the manual:

a) Charging the Capacitor with a Resistorb) Discharging the Capacitor with a Resistor

In the report, do include the following:a) Circuit diagram, Circuit ExplanaJonb) Final circuit picture c) Pictures of MulJmeter readings: Step No. 8, Step No. 9, Step No. 10 and Step No. 11d) ObservaJone) Reasoning (if any)f) Result (if any)g) General Theory: About components used (LED, resistor, mulJmeter, DC baHery, capacitor).

Project Report

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 rst 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.

TroubleshooLng Tips

Result

PrecauLon

Notes

Circuit

Measure

AcLvity

Advantages

ModicaLon

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