Procedure: D) Discharging a capacitor through a resistor. If one connects a char

Question

Procedure:

D) Discharging a capacitor through a resistor. If one connects a charged capacitor to a resistor, the charge stored on the capacitor will flow out. The energy stored in the capacitor is converted into heat in the resistor. The equation that governs this process is another example of exponential decay:Q(t)=Q0e(–t / RC) (4) Because V=CQ, the voltage will also follow an exponential decay:

V(t)=V0e(– t / RC)

(5) In equations ( 4) and (5) the quantity RC has units of seconds, and is called the time constant of the circuit.Connect the circuit as shown in Fig. 2 using the breadboard, the 1F capacitor and a 1M resistor.

Connect the MinigrabbersTM from the voltage follower to the 1 F capacitor. Open the experiment fileas directed on the whiteboard. Charge the 1 F capacitor from the 5V power supply on the breadboard unit. Click on the “Record” button (bottom left) in the experiment file and disconnect the power supply from the circuit. The programis set to start recording the data when the voltage across the capacitor drops below 4.5 V, and stops automatically after 5s. Data is automatically fitted to exponential curve.

The program fits the data to a function: y = Ae -Bt + y0, where the coefficient B will represent (1/Time Constant). Calculate the experimental time constant for your RC circuit and compare it with the expected deduced from the known values of the circuit components (can disregard this, need unknown value found). Replace the 1µF capacitor with the unknown capacitor used in step 4, part C of the lab procedure. Measure the time constant as described above and find the value of the unknown capacitor. I have provided the graph and experimental data

Evaluating the Unknown (2 trials)

Capacitance of circuit (µF)

Capacitor Used to Charge (µF)

Voltage (V)

1.1

0.839

2.164

1.1

0.755

1.852

Capacitance of circuit (µF)

Capacitor Used to Charge (µF)

Voltage (V)

1.1

0.839

2.164

1.1

0.755

1.852

capacitors (Last saved by user) -Word FILE HOME INSERT DESIGN PAGE LAYOUT REFERENCES MAILINGS REVIEW VIEW DEVELOPER s Cut Garret Heltermes Calibri (Body) Copy Paste Find c Replace Select Editing Blu-abex: x1 A-·. .-R- Format Painter 1 Normal 1 No Spac Heading 1 Heading 2 Clipboard Title Subtitle Subtle Em Emphasis Font Paragraph Tabie tide here) Run #1 valtage 4.972 0.300 0.350 3.724 3.219 2.994 2.784 0 500 2.407 2.238 2081 934 1.799 1 573 Natural Exponennal 25 0.700 0.750 0.800 0.850 00010 35x104 A=619 1 45 RMSE 0.00126 .000 .447 1.150 1 200 1.082 1.00 0.936 0.870 1.300 1.350 1.400 LO 1.5 20 2530 35 4.045 505 0 70 Time (s SECTION1 PAGE 4 OF6 511 WORDS + 26056 713 AM 9/18/2017

Explanation / Answer

Time constant is equal to the time, when charge or potential drops to 1/e times the initial value of charge or potential.

As initial potential is 5V,

At one time constant, the value of voltage will be

V(tau) = 5/e V = 1.83939 V

And from the given graph, we can see that potential falls to 1.8 around in 0.9 seconds.

Hence, value of time constant RC = 0.9 seconds

Given resistance = 1 Mega Ohm

Hence , if

RC = 0.9

C = 0.9/R = 0.9/(106)

C = 0.9 x 10-6 F

Or

C = 0.9 micro Farad.

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