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Date___________

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CAPACITORS EXPERIMENT

Short description:

In this experiment you will determine how voltages are distributed in capacitor circuits and explore series and parallel combinations of capacitors.

Equipment:

−

−

−

−

AC power supply (set to 10V)

Three 0.1µF capacitors, one 0.01µF capacitor, one unknown capacitor

Multimeter

4 cables

Theory:

Capacitors are electronic devices which have fixed values of capacitance and negligible resistance. The capacitance C is the charge stored in the device, Q, divided by the voltage difference across the device, ∆V:

C = Q/∆V

(1)

The schematic symbol of a capacitor is has two vertical (or horizontal) lines a small distance apart (representing the capacitor plates) connected to two lines representing the connecting wires or leads).

There are two ways to connect capacitors in an electronic circuit - series or parallel connection.

Series:

In a series connection the components are connected at a single point, end to end as shown below: C1

C2

For a series connection, the charge on each capacitor will be the same and the voltage drops will add. We can find the equivalent capacitance, Ceq, from

Q · 1/Ceq = ∆V = ∆V1 + ∆V2 = Q/C1 + Q/C2 = Q [1/C1+1/C2]

1/Ceq = 1/C1 + 1/C2

(2)

(3)

Parallel:

In the parallel connection, the components are connected together at both ends as shown below:

C1

C2

For a parallel connection, the voltage drops will be the same, but the charges will add. Then the equivalent capacitance can be calculated by adding the charges:

Ceq ∆V = Q = Q1 + Q2 = C1∆V + C2∆V = [C1+C2] ∆V

Ceq = C1 + C2

(5)

(6)

Procedure:

Figure 2. AC power supply

Figure 1. Multimeter

1. Turn on the power supply and set the AC voltage to 10 V. Measure the actual power supply voltage with the multimeter and record it below:

VPS = ___________________V

2. Connect two 0.1 µF capacitors in series. Measure V2

(across C2) and record it below.

V2 (measured) = ____________ V

3. Compute the expected value of V2 using VPS, the values of C1 and C2 with equations 1 and 3.

Remember that Eq. 1 is true for each capacitor, including the combined C12.

C2

C1

V

V2 (expected) = ____________ V

% difference = |measured - expected| / measured x 100 % = __________

4. Connect a third 0.1 µF capacitor in parallel with C2.

Compute their equivalent capacitance C23.

∼

C2

C23 = _________ µF.

Measure and compute the voltage across C2. [Hint: is this the same as the voltage across the equivalent capacitor C23? You may want to compute the total equivalent capacitance seen by the power supply, C123]

C3

V2 (measured) = ________ V,

V

V2 (expected) = ________ V,

% difference = ________

5. Now remove the third capacitor C3 and replace it with a 0.01 µF capacitor.

Compute their equivalent capacitance C23.

C23 = _________ µF.

Measure and compute the voltage across C2

V2 (measured) = ________ V,

V2 (expected) = ________ V,

% difference = ________

C1

∼

6. Now connect the 0.1 µF and the 0.01 µF capacitor in series. as C2 and C3. Compute the equivalent capacitance C23

C2

C3

C1

C23 = _________ µF.

V

Measure and compute the voltage across the equivalent capacitance C23.

V23 (measured) = ________ V,

V23 (expected) = ________ V,

% difference = ________

7. This method can be used to find an unknown capacitance. Replace C2 with the unknown value capacitor and determine its capacitance by measuring V2 and using equations 1 and 3.

V2 = ___________ V,

C 2 = _________ µF.

∼

C2 = ?

V

C1…...

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