Answer DIscussion Questions 1-5 Based on the following data: Capacitors in Circu
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Question
Answer DIscussion Questions 1-5 Based on the following data:
Capacitors in Circuits
EQUIPMENT NEEDED:
– AC/DC Electronics Lab Board: Capacitors, Resistors, Wire Leads – D-cell Battery –Digital Multimeter – Stopwatch or timer with 0.1 sec resolution.
Purpose: The purpose of this lab will be to determine how capacitors behave in R-C circuits. The manner in which capacitors combine will also be studied.
Experiment data: Your circuit board contains a battery, 2 resistors and 2 capacitors.
Use the color code picture below to identify the resistors:
2. Record in Table 1 the resistance of the resistors by the color code and due to color confusion, it would be wise to measure the resistance using a multimeter. This time, use the multimeter across each resistor, with the dial on the resistor scale (M) and record in Table 1. Do not forget to convert from M to k (multiply by 1000).
Table 1.
Resistor
Color Code
Resistance by color code (k)
Resistance by using multimeter (k)
Resistor 1
Red, red, yellow
220
210
Resistor 2
Brown, black, yellow
100
99100
3. Now take a look at the two capacitors placed on the circuit board. Read the value
written on them and record in Table 2.
Capacitor
Capacitance (µF)
Capacitor 1
47
Capacitor 2
100
Time constant is defined by the product between resistance and capacitor and it is measured in seconds.
4. Use =RC and your knowledge of units for both resistor and capacitor, to prove that is measured in seconds. HINT: show how units cancels out and what is left is seconds.
Show your work below:
5. For Table 3, despite the fact that resistors and capacitor are given in k and F, when you calculate the time constant you have to convert in
and F. Table 3
Resistor R (k)
Capacitor (F)
=RC (s)
Resistor 1= 220
Capacitor 1=47
10.34
Resistor 2= 100
Capacitor 1= 47
4.7
Resistor 1= 220
Capacitor 2= 100
22
Resistor 2= 100
Capacitor 2= 100
10
Measure the voltage at the battery using the digital multimeter. Record the value below:6. Measured voltage of the battery = 1.3V
The time constant =RC is the amount of time it takes for the capacitor to store 63% of the voltage.
7. Based on this fact, calculate at what voltage reading will you reach one time constant?
Voltage after one time constant V=.819 V
Figure 1
1. Connect the circuit shown in Figure 1, using one battery, the switch, a resistor and a capacitor. All the components are already on the board, use one resistor and capacitor only.
2. Connect the Digital Multimeter so the black “ground” lead is on the side of the
Lab experiment procedure:
capacitor that connects to the negative terminal of the battery and set it so that it reads up to a maximum of 1.5 V DC (max voltage at the battery).
3. Start with no voltage on the capacitor and the switch off. If there is remaining voltage on the capacitor, use a piece of wire to touch the ends of the wire to points B and C as shown in Figure 1 to discharge the capacitor.
4. Now close the switch by pushing and holding the button down. Observe the voltage readings on the Multimeter, the voltage across the capacitor.
5. If you now open the switch by releasing the button, the capacitor should remain at its present voltage with a very slow drop over time. This indicates that the charge you placed on the capacitor has no way to move back to neutralize the excess charges on the two plates.
6. Connect a wire between points A and C in the circuit, allowing the charge to drain back through the resistor. Observe the voltage readings on the Multimeter as the charge flows back.
7. Now repeat steps 3-5, this time recording the time taken to move from 0.0 volts to VOLTAGE after one time constant (calculated previously at 7.) while charging, tC, and the time taken to move from the maximum voltage from the battery (consider highest value provided by the battery, it might not be 1.5 volts because the battery has been used by now) to 0.45 volts while discharging, tD.
8. Record your times along with the resistance of resistor 1 and capacitance of capacitor 1 values in Table 4.
9. Replace the capacitor 1 with capacitor 2 (you got two different capacitors on the board), keep the same resistor 1 and Repeat step 7, recording the charging and discharging times in Table 4.
10. Return to the capacitor 1, but change the resistor, now use resistor 2 in the circuit. Repeat step 7, recording your data in Table 4.
Trial
Resistance (k)
Capacitance (µF)
TC(s)
T0(s)
T=RC
1
220
47
23.7
12.30
10.34
2
220
100
21.36
25.18
22
2
100
47
6.86
8.65
4.7
11. Return to the resistor 1, but use the capacitor 1 in series with the capacitor 2. Repeat step 7, recording your results in Table 5.
R= 220 K C1= 100 µF C2=47µF
Calculate the equivalent capacitate using : 1/ceq=1/c1+1/c2
Ceq=0.031 (µF) record in table 5.
12.Now repeat step 7, but with the capacitors in parallel. Calculate the equivalent capacitate using: Ceq=C1+C2
Ceq= 147(F) Record in Table 5.
Types of circuit
R (k)
Ceq(µF)
T(s)
Tc(s)
To(s)
Series
220
0.031
6.8x103
9.86
7.71
Parralel
220
147
10.34
35.30
33.45
Table 5:
Discussion:
How would you describe the manner in which the voltage changes? It would be reasonable to sketch a graph showing the manner in which the voltage rose over time.
How would you describe the manner in which the voltage falls? (It would be reasonable to sketch a graph showing the manner in which the voltage fell over time.)
What is the effect on charging and discharging times if the 2 capacitors are in series? How did it change compared with one capacitor in circuit? Why did it change? Discuss your finding.
4. What is the effect on charging and discharging times if the 2 capacitors are in parallel? How did it change compared with one capacitor in circuit? Why did it change? Discuss your finding.
5. What is the effect on charging and discharging times if the resistance of the circuit is increased? What mathematical relationship exists between your times and the resistance?
Resistor
Color Code
Resistance by color code (k)
Resistance by using multimeter (k)
Resistor 1
Red, red, yellow
220
210
Resistor 2
Brown, black, yellow
100
99100
Explanation / Answer
1.. Variation of Volatge In RC ckt while charging is given by
V = Vo (1- e^-t/Rc)
Where RC is the time Constant of the ckt and
Vo is the initial Volatge
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2. Variation of Volatge In RC ckt while charging is given by
V = Vo (e^-t/Rc)
Where RC is the time Constant of the ckt and
Vo is the initial Volatge
--------------------------------------
4. When Two Capacitors are connected in series,
Cnet = 1/Cs = 1/C1 + 1?c2
so
net capaciatnce C decreases as they are in parallel
also as Q = CV , charge also decreases
---------------------------------------------
5. Now again if R's are in series, RNet increases
RC time consatnt increases there by from the equation
V = V (1-e^-t/RC)
V decrases, as Q = CV
Q = charge = decreases
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