Costo estimado del plan de calidad

Worked Through Examples

1. Find the time constant of a circuit containing a 10-kilohm resistor in series with a 0.82-microfarad capacitor.

To solve this problem, you must use the time constant formula T = RC. Substituting in the circuit values, the formula reads T = 10 k X 0.82 F. In scientific notation the values are: T = 1.0 X 104 _{X 8.2 X 10}-7.

_{1.0 X 10}4

X 8.2 X 10-7

T = 8.2 X 10-3 _{seconds (s) or 8.2 milliseconds (ms)}

2. Find the time constant of this circuit:

Use the formula: T = RC. First substitute in the circuit values: R = 100 k , C = 20 F.

T = 100 k , C = 20 F T = 1.0 X 105 _{X 2.0 X 10}-5

3. How long will it take the capacitor in the following circuit to reach full charge?

First, use the time constant formula T = RC T = RC

T = 8.2 M X 560 F T = 8.2 X 106 _{X 5.6 X 10}10

T = 4.59 X 10-3 _{s or 4.59 ms}

You must remember that the RC time constant formula you just worked gives you one time constant (in seconds). Five time constraints are required for full charge. So, multiply the time constant by 5 to arrive at the correct answer.

4.59 X 10-3

X 5

22.95 X 10-3 _{or 2.3 X 10}-2 _seconds

The capacitor will be fully charged after 2.3 X 10-2 _{seconds or 23}

milliseconds.

4. Find the voltage across the capacitor in the circuit shown below 500 milliseconds after the switch is closed. (Use the universal time constant graph.)

First, you should calculate the time constant of the circuit. T = RC T = RC T = 10 k , X 33 F T = 1.0 X 104 _{X 3.3 X 10}-5 T = 3.3 X 10-1 _{or 330 ms}

Now look at the universal time constant graph. Time

(horizontal axis) is measured in time constants. To convert this chart to seconds, multiply 330 milliseconds by each of the time divisions. For example:

1 X 330 ms = 330 ms 1.5 X 330 ms = 495 ms 2 X 330 ms = 660 ms 3 X 330 ms = 990 ms 4 X 330 ms = 1.32 s 5 X 330 ms = 1.65 s

Now these values are applied to the universal time constant graph.

Look at the chart and locate the 500 millisecond position on the horizontal axis. Now trace directly upward (following the dotted line) and note the point on the charging curve that is reached at 500 ms.

Tracing to the left from that point, across the graph, you can see that the amplitude at the intersection point is about 0.78 or 78% of the full charge voltage; 0.78 X 100 V. So after 500 ms. the capacitor is charged to 78 volts.

5. Find the charge in coulombs of the capacitor in problem 4, at the end of 500 milliseconds.

The formula for calculating the charge stored in a capacitor is Q = CE

where

Q = the stored charge in coulombs C = the capacitance in farads

E = the voltage between the capacitor plates Substituting the values of capacitance and voltage:

Q = 33 F X 78 V Q = 3.3 X 10-5 _{X 7.8 X 10}1

Q = 2.57 X 10-3 _{coulombs (or 2.57 millicoulombs)}

6. Using the universal time constant graph, calculate the time required for the capacitor shown below to charge to 55 volts.

First, calculate the circuit’s time constant using the formula: T = RC

T = RC T = 470 k , X 18 F T = 4.7 X 105 _{X 1.8 X 10}-5

Now, the universal time constant curve may be used as follows in solving this problem. First, examine the vertical axis. On this axis the fraction of the maximum voltage is located. The

maximum voltage here is 120 volts: the total applied voltage. What fraction of 120 volts is 55 volts? Thus, 55/120 equals 0.458. This is the fraction of the applied voltage 55 volts represents. Now, locate 0.458 on the vertical axis of the universal time constant graph. Trace to the right horizontally (a dotted line is drawn in for you to follow) until you intersect the charging curve.

Locate that point on the curve, and then trace directly down to the horizontal axis. At this point you read the time elapsed: 0.6

time constants. You know that 1 time constant is 8.46 seconds, so the total elapsed time is 0.6 X 8.46 or 5.08 seconds.

7. A “strobe” flash attachment for a camera has a bulb that requires 0.02 coulomb of charge at 450 volts in order to flash properly. What is the minimum size capacitor that could be satisfactorily used?

Since both the quantity of charge (Q) and voltage (E) are known, the equation C = Q/E can be used to solve this problem. Simply substitute in the capacitor values and solve for C.

C = Q/E C = 0 .02 C

450 V

(coulomb )

8. Find the approximate frequency of oscillation in the circuit shown here.

The circuit shown above is a “relaxation oscillator.” It operates on the basis of its RC time constant. The bulb shown connected across the capacitor is an NE-2 neon glow lamp. These lamps require a certain voltage (called the “firing voltage”) in order to light. Once lit, the voltage across the lamp must fall significantly below the firing voltage before it will turn “off.” Typical “on” and “off” voltages for neon glow lamps are: 75 volts “on” and 50 volts “off.” This means that the typical NE-2 will not “light” until the voltage across it reaches 75 volts, but once lit, will continue to glow until the voltage drops below 50 volts. Before the lamp lights, it has a very high resistance (essentially an open circuit). Once the lamp is on, its resistance drops to a low value. Consider what will happen when one of these lamps is

connected across a capacitor as shown in the circuit above. When power is applied to the circuit, the capacitor will begin to charge up to the source voltage. The rate of charging will be controlled by the RC time constant. When the capacitor reaches 75 volts, the neon bulb (which is connected in parallel with the capacitor) will also have 75 volts applied across it. At this instant, the bulb will light, allowing heavy current flow, and thus discharging the capacitor very quickly. As the capacitor discharges, its voltage will drop down below the 50 volts required to keep the neon bulb lit. The bulb goes out and the capacitor again charges up to the 75 volts required to fire the bulb, and the cycle is repeated again and again. As you can see, there are several factors that affect the rate of blinking (or

oscillation) of the bulb: the resistor size, the size of the capacitor, the supply voltage, and the characteristics of the individual neon bulb.

To analyze this problem, first calculate the RC time constant of the circuit and plot it on a universal time constant graph.

T = RC T = 7.5 M , X 10.2 F T = 7.5 X 106 _{X 2.0 X 10}-7 T = 1.5 s 1 X 1.5 s = 1.5 s 1.5 X 1.5 s = 2.25 s 2 X 1.5 s = 3.0 s 3 X 1.5 s = 4.5 s 4 X 1.5 s = 6.0 s 5 X 1.5 s = 7.5 s

To give a clearer picture of the operation of this circuit, these values are plotted on the horizontal axis of the universal time constant graph above.

The lamp fires at 75 volts, and causes the voltage across the capacitor to rapidly drop to 50 volts so that the lamp then goes out. Voltage across the capacitor, plotted as time goes on, will appear as shown on the next page.

In order to find the time duration between flashes, simply look back at the Universal Time Constant graph you just filled in. Locate 75 volts and 50 volts, and measure the time elapsed between these two points. Seventy-five volts occurs at approximately 1.4 time constants or 2.1 seconds. Fifty volts occurs at 0.7 time constants or 1.055 seconds. The time elapsed is the difference between the two times. Subtract and you get 2.1 s - 1.05 s = 1.05 s. So the lamp will blink once every 1.05 seconds. Dividing 60 by 1.05 yields a frequency of 57 flashes per minute. 9. Calculate the total capacitance of this circuit.

Problems of the type shown above give many students

headaches because capacitors “add” just the opposite of the way resistors do. Parallel capacitors are added by using a formula similar to the series resistance formula: CT = C1 + C2 + C3 ...

Series capacitors must be added by using a formula similar to the parallel resistance formula:

CT = 1

1 /C₁ 1 /C₂ 1C_{3 . . .}

To solve this problem, the 4-microfarad and the 6-microfarad capacitors should be combined by using the parallel capacitance formula: CT = C1 + C2 + C3 ...

CT = 4 F + 6 F

The 10 microfarads of capacitance must be combined with the 8 microfarads of capacitance by using the series capacitance formula. CT = 1 1 /C₁ 1 /C₂ 1C_{3 . . .} CT = 1 1 / 10 1 / 8 CT = 1 0 .1 0 .125 CT = 1 0 .225 CT = 4.44 F

10. Calculate the total capacitance of the following circuit.

First, find the total capacitance of the upper circuit branch using the series capacitance formula:

CT = 1 1 /C₁ 1 /C₂ 1C_{3 . . .} CT = 1 1 / 4 1 / 8 CT = 1 0 .25 0 .125 CT = 1 0 .375 CT = 2.67 F

Now the total capacitance may be found by combining the two parallel capacitances using the parallel capacitance formula CT =

C1 + C2 + C3 ...

CT = 2.67 F + 6 F

Practice Problems

Depending upon the approach you use in solving these problems and how you round off intermediate results, your answers may vary slightly from those given here. However, any differences you may encounter should only occur in the third significant digit of your answer. If the first two significant digits of your answers do not agree with those given here, recheck your calculations.

d.

e.

Answers 1.a. T = 7.05s 1.b. T = 165 ms 1.c. T = 4.73 ms 1.d. 560 s 1.e. T = 5.6 s 2.a. CT = 1 F 2.b. CT = 1.33 F 2.c. CT = 5 F 2.d. CT = 0.75 F 2.e. CT = 10 F 3.a. Q = 900 C 3.b. Q = 3.6 mC 3.c. C = 390 F 3.d. E = 1.33 V 3.e. Q = 72 C 4.a. T = 2.2s 4.b. 11 s 4.c. 39.5 V 4.d. 75.8 V 4.e. 1.03 s

Inductors and the