Circuit Current Related to a Capacitor’s Charg

In the circuit of Figure 10.1, there is a 10 kilohm resistor in series with a 1 microfarad capacitor connected through a switch, S1, to a 10-volt dc power supply. The switch is open and the capacitor, therefore, has no charge on its plates.

Recall the basic action of a capacitor in this type of circuit was discussed in Chapter 6. When the switch first contacts position 1, at time t₁, as shown in Figure 10.2, the voltage is applied across the plates of the capacitor. Electrons then move from the upper plate of the capacitor through the circuit to the lower plate. This flow of electrons constitutes a current which will be called the charge current.

Calculation of I_max

The initial current is large, but its maximum value is limited by the value of the resistance in the circuit. The maximum value of the charge current is equal to the applied voltage, E_T, divided by the value of the resistance.

(10–1)

In the circuit of Figure 10.1, E_T = 10 volts and R = 10 kilohms. Therefore,

10 volts divided by 10 kilohms equals 1 milliampere of current.

As more and more electrons are transferred and accumulate on the negative plate, it becomes increasingly difficult for the applied voltage to transfer additional electrons and the transfer rate decreases. Thus, the current in the circuit decreases. This current decreases until it is finally zero at which time the capacitor is fully charged.

A graphic representation of this action is shown in Figure 10.3, where the current is plotted versus time. At first, at time t₁ the current is at its largest value, 1 milliampere, then decreases to zero exponentially as the capacitor charges.

Variation of E_C

As this transfer of electrons occurs, as shown in Figure 10.4, current flows in the circuit, the charge on the capacitor builds, and a difference of potential is present across its plates. When the capacitor is fully charged, the voltage across it, E_C, is equal to the applied voltage of 10 volts.

Figure 10.5 is a plot of that voltage, E_C. Initially, the voltage across the capacitor is zero, but as the electrons are transferred, beginning at time t₁, the capacitor charges exponentially to the applied voltage of 10 volts during a period of time. Compare Figures 10.3 and 10.5 and note that the voltage across the capacitor increases as the current flow decreases and the capacitor charges.

Calculation of E_R

The voltage across the resistor depends upon two factors: 1) the amount of current flowing through it, and 2) the value of the resistor. More specifically, by Ohm’s law,

(10–2)

Since the value of the resistor is constant (10 kilohms), the voltage across it depends upon the amount of current flowing in the circuit at any particular time. When the switch is first closed, as shown in Figur 10.6, the voltage across the resistor is equal to:

So at t₁, the voltage across the resistor is 10 volts.

When the current is zero, the voltage across the resistor is equal to zero because:

This is illustrated in Figure 10.7.

Graphically, the voltage across the resistor would appear as shown in Figure 10.8. The voltage is 10 volts when the switch is closed, and then it decreases exponentially to zero as the current decreases and as the capacitor charges.

Remember that even if the switch is open and the dc source removed the capacitor will retain its charge.

Change in the Direction of Current Flow

When S1 is placed in position 2, the capacitor has been provided with a discharge path through the resistor. Notice in Figure 10.9 that the direction of electron flow (current) when discharging is opposite to the direction of the current while charging.

The current is maximum and decreases to zero exponentially as the electrons transfer. This action is shown graphically in Figure 10.10. Since the direction of flow is opposite, the current values are plotted as negative values on the graph. This illustrates graphically that the discharge current direction is opposite to the charge current direction. The charge current direction was arbitrarily chosen as positive.

Calculation of I_max

The maximum amount of current during discharge depends upon the voltage to which the capacitor is charged and the value of the discharge path resistance. In this case, the resistance is the same as it was for the charge path.

The maximum discharge current is calculated:

In the example circuit, E_C = 10 volts and R = 10 kilohms. Therefore,

1 milliampere is the discharge current. The discharge current decreases from the Imax value exponentially as the capacitor discharges. When the charge on the plates is balanced, no more electrons are transferred, the voltage across the capacitor is zero, and the circuit current is zero.

Variation of E_C

As the capacitor discharges, the voltage across the capacitor is 10 volts initially at t₂, and then decreases to zero exponentially, as shown graphically in Figure 10.11.

Calculation of E_R

The voltage across the resistor during capacitor discharge depends upon the amount of current passing through it at any particular time, as it did during capacitor charge. However, since the current during discharge flows in a direction opposite to its direction during the charging process, the polarity of the voltage across the resistor will be opposite to the polarity it had during charging as shown in Figure 10.12.

This opposite polarity is shown by indicating the values of voltage across the resistor during discharge as being negative as shown in Figure 10.13. Remember that the voltage values were arbitrarily chosen as being positive because the charging current direction was chosen as positive for the charge process.

The maximum voltage across the resistor during discharge is equal to the voltage to which the capacitor is charged.

(10–3)

In this case, that voltage is 10 volts. This is evident from the fact that the voltage across the resistor can be written, by Ohm’s law, as being equal to the current through it times its value of resistance: E_R = IR. Therefore, in the circuit,

The voltage across the resistor is equal to 10 volts when the switch is initially placed in position 2. The voltage across the resistor directly dependent on the circuit current flowing, decreases to zero exponentially as the current decreases to zero. This shown in Figure 10.13. Note the shape is the same as the current waveform of Figure 10.10.

RC Time Constant Defined

The time required for a capacitor to fully charge is measured in terms of a quantity called the time constant of the circuit.

The time constant of an RC circuit is defined as the amount of time necessary for the capacitor to charge to 63.2 percent of its final voltage.

The time constant is symbolized by the Greek letter, tau, τ.

For an RC circuit, tau is equal to the value of the resistance times the value of capacitance in the circuit. Expressed mathematically,

(10–4)

Where tau is measured in seconds, R in ohms, and C in farads.

Calculation of an RC Time Constant

The resistance of the example RC circuit being discussed (Figure 10.1) is 10 kilohms. Its capacitance is 1 microfarad. The time constant for the circuit is calculated:

The time constant is 10 milliseconds.

In the circuit, the capacitor is charging to 10 volts. 63.2 percent of 10 volts is 6.32 volts. Thus, as shown in Figure 10.15, it takes 10 milliseconds or one time constant of time for the capacitor to charge to 6.32 volts.

In the next time duration of one time constant the capacitor will charge to 63.2 percent of the remaining value of voltage between the maximum, 10 volts, and its present charge, 6.32 volts. Therefore,

and

Thus, at the end of two time constants or after 20 milliseconds of charging, as shown in Figure 10.16, the capacitor will have charged to

This process continues with the capacitor charging 63.2 percent of the remaining difference during each time constant of time until after about five time constants, the capacitor, for all practical purposes, is fully charged with a charge of 9.93 volts as shown in Figure 10.17. Thus, it is assumed for practical cases that it takes five time constants for the capacitor to fully charge.

In the example circuit with a time constant of 10 milliseconds, the time required for the capacitor to fully charge to 10 volts can be calculated:

It will require 50 milliseconds for the capacitor to fully charge.

UTCC Related to E_C

Note that the increasing curve of Figure 10.19 has a shape exactly like the shape of the rise of voltage across the capacitor. This can easily be proved true by plotting the voltages across the capacitor in the circuit of Figure 10.1 against time measured in time constants for exact multiples of one time constant.

Initially, the voltage across the capacitor is zero when the switch is closed. The origin represents time equals zero and voltage equals zero. This is the time when the switch closes with no initial voltage across the capacitor. As calculated previously and shown in Figure 10.17, after one time constant (10 milliseconds), the voltage across the capacitor equals 6.32 volts (63.2 percent of maximum).

After two time constants, 20 milliseconds, the voltage across the capacitor equals 8.65 volts or 86.5 percent of maximum. After three time constants, 30 milliseconds, E_C equals 9.5 volts or 95 percent of maximum. After four time constants, 40 milliseconds, E_C equals 9.82 volts or 98 percent of maximum. And after five time constants, the voltage across the capacitor is equal to 9.93 volts or 99 percent of maximum.

When the capacitor has reached 99 percent of maximum it is virtually fully charged. At this point the capacitor is considered to be at 100% of applied voltage even though the actual charge is 99% of applied voltage. The error is negligible.

If all of the points given above are connected with a smooth curve, the result is the rising curve shown on the UTCC of Figure 10.18 and Figure 10.19.

Exponential Characteristic of the UTCC

The shape of these curves follows a precise mathematical relationship which involves an exponent containing the time constant term. Because of the exponent containing the time constant term, they are called exponential curves. This characteristic results in the terms exponential increase or exponential decrease. Because these exponential equations are complex mathematically, the chart is used to provide a direct-read capability.

UTCC Related to Circuit Current and E_R

Recall that it was stated earlier that the exponential curve applies to decreasing waveforms as well as to increasing waveforms. An example is the circuit current waveform which is maximum initially and decreases to zero in five time constants, as shown by the curve in Figure 10.20. The same waveform applies to the voltage across the resistor. This decaying waveform is simply a mirror image of the rising exponential waveform shown in Figure 10.19.

At one time constant the decaying waveform has reduced 63.2 percent from its initial value. At two time constants the waveform has reduced 86.5 percent, 95 percent at three time constants, 98 percent at four time constants, and at five time constants an assumed reduction of 100 percent.

This may be expressed another way by changing the scale of the curve to percent of initial value as shown in Figure 10.21. In one time constant the exponential curve has decreased by 63.2 percent, to 36.8 percent of its initial value. It is down to 13.5 percent of its initial value in two time constants, 2 percent in four time constants, and it is fully decayed in five time constants.

Using the UTCC

Any instantaneous voltage or current value across or through the components of a dc resistive-capacitive circuit during the charge and discharge of the capacitor can be determined by using the Universal Time Constant Chart. (Later in this chapter, it will also be used to predict values of current and voltage while current is increasing and decreasing in dc resistive-inductive circuits.) For instance, using the UTCC the percent of full charge on the capacitor after any charging time or time constant period can be determined.

Figure 10.22 shows that after one and one-half time constants the percent of full charge, from the UTCC chart, is approximately 78 percent. If the applied voltage is 10 volts, as shown in Figure 10.23a, the voltage across the capacitor, E_C at this time is 78 percent of 10 volts which is 7.8 volts. This is shown graphically in Figure 10.23b. If 50 volts were applied, as shown in the circuit of Figure 10.24a, E_C is 78 percent of 50 volts which is 39 volts. This is shown graphically in Figure 10.24b.

If, however, one wants to know the current in the circuit or the voltage across the resistor after a specified number of time constants the percent of maximum must be read from the decaying exponential waveform. After one and one-half time constants, for example, the percent of maximum read from the chart for current or voltage across the resistor is 22 percent as shown in Figure 10.25. If the maximum current in the circuit is 10 milliamperes then after one and one-half time constants, the current has decreased to 22 percent of 10 milliamperes which is 2.2 milliamperes (0.22 × 10 mA = 2.2 mA). This shown graphically in Figure 10.26.

If the maximum voltage across the resistor was 30 volts, then after one and one-half time constants, the voltage across the resistor has decreased to 22 percent of 30 volts which is 6.6 volts (0.22 × 30 V = 6.6 V). This is shown graphically in Figure 10.27.

During the discharge of the capacitor, the chart again is used. However, the appropriate curve must be chosen.

Converting Time to Time Constants

Sometimes it is necessary to convert from time to time constants. The technique is as follows. To find the number of time constants that a specified time represents, simply divide the time, in seconds, by the number of seconds in one time constant.

(10–5)

For example, if the value of R in an RC circuit is 100 ohms and the value of C is 200 microfarads, then the time of one time constant, τ, is equal to R times C, and

In this example, τ equals 20 milliseconds.

If the number of time constants represented by 60 milliseconds is desired, this is calculated by dividing the time in question, 60 milliseconds in this case, by the length of one time constant, 20 milliseconds.

In this example, there are three time constants in 60 milliseconds.

For the same R and C, the number of time constants in 4 milliseconds is calculated:

In this case, there is 0.2 of a time constant in 4 milliseconds. Similar problems are solved by performing similar calculations.

Solving for Instantaneous Voltage and Current Values: Charging

The original circuit, shown in Figure 10.1 and repeated in Figure 10.28, consists of a 10 kilohm resistor, a 1 microfarad capacitor, and a 10-volt dc source. Recall that one time constant for this circuit equals 10 milliseconds. The voltage across the capacitor, the voltage across the resistor and the current after the capacitor has been charging for 5 milliseconds (5 milliseconds after the switch contacts position 1) can be calculated using the information about time constants discussed in the previous section.

5 milliseconds in terms of time constants is calculated:

In the example circuit of Figure 10.28, 5 milliseconds equals 0.5 time constants. The waveforms for the circuit are as shown in Figure 10.29. The maximum capacitor charge, EC, is 10 volts, or the applied voltage. The maximum current initially, is 10 volts ÷ 10 kilohms = 1 milliampere. The maximum voltage across the resistor is 10 volts, which is the same as the capacitor’s maximum charge.

The Universal Time Constant Chart of Figure 10.20 can be used to determine the percent of maximum voltage across the capacitor. The rising curve, repeated in Figure 10.30, shows that the percent of maximum voltage across the capacitor after one-half of a time constant is about 40 percent.

Thus, the voltage across the capacitor after one-half of a time constant (5 milliseconds) is 4 volts.

Using the decaying curve as shown in Figure 10.31, the percent of maximum current in the circuit after one-half of one time constant is determined to be 60 percent. Said another way, the current has decayed 40 percent from its initial value to a 60 percent value.

Thus, the current in the circuit after 5 milliseconds is 0.6 milliamperes.

The voltage across the resistor in the example RC circuit is equal to 60 percent of its maximum value of 10 volts, which is 6 volts (0.60 × 10 = 6).

The voltage across the resistor could also have been calculated using Ohm’s law:

Solving for Instantaneous Voltage and Current Values: Discharging

When the switch is placed in position 2 as shown in Figure 10.32 (Figure 10.9 repeated) the capacitor begins to discharge. This changes circuit action. The instantaneous voltage across the capacitor, across the resistor, and the current in the circuit at a selected time after the switch contacts position 2 can be calculated. This is best described using an example. The values calculated will be those present 30 milliseconds after the switch contacts position 2.

The number of time constants represented by 30 milliseconds is calculated:

In this case, there are three time constants. Waveforms for the circuit are as shown in Figure 10.33. The maximum capacitor voltage, EC, is 10 volts at the instant that it begins discharge. The maximum current, I_max is equal to the charge on the capacitor as it begins discharge divided by the resistance.

(10–6)

In this example, I_max = 1 milliampere. Remember, however, that the current is flowing opposite to the direction of current flow during charge. Thus, the maximum discharge current is designated as a negative 1 milliampere.

The voltage across the resistor during discharge is equal to the voltage across the capacitor. The voltage is specified as negative during discharge since it is opposite in polarity to the voltage which appeared across the resistor during the charge of the capacitor.From the Universal Time Constant Chart in Figure 10.34, using the decaying curve, the percent of maximum voltage or current after three time constants is determined to be four percent. Thus, after three time constants:

The charge on the capacitor has decreased from 10 volts to 0.4 volts after three time constants.

After three constants, the current will decrease from I_max to 4 percent of I_max:

Therefore, after three time constants I = −40 microamperes.

After three time constants, the voltage across the resistor will decrease from E_Rmax to 4 percent of E_Rmax.

Thus, E_R = −0.4 V.

RC Time Constants Related to Kirchhoff’s Voltage Law

It is interesting to note that Kirchhoff’s voltage law is valid for dc RC circuits. At any instant in time the voltage across the resistor plus the voltage across the capacitor equals the applied voltage: E_R + E_C = E_A. Initially, at t₁, when the switch in the circuit of Figure 10.28 is first contacted at position 1, the voltage across the resistor equals 10 volts, or the applied voltage, and the voltage across the capacitor equals zero volts. This is shown in Figure 10.35. The sum of these two voltages equals 10 volts, or the applied voltage.

After one-half time constant, as calculated previously, the voltage across the resistor is 6 volts and the voltage across the capacitor is 4 volts. The sum of these voltages is 10 volts (6 V + 4 V = 10 V).

After five time constants when the voltage across the resistor is 0 volts, the capacitor has charged to 10 volts. The sum of these voltage drops, too, is 10 volts (0 V + 10 V = 10 V).

Analysis of Circuit Current

The dc RL circuit shown in Figure 10.36 includes a 1 kilohm resistor in series with a 10 millihenry inductor connected through switch S1 to a 10 volt dc power supply. When the switch is open, the current in the circuit is zero.

Remember the basic action of an inductor in the type of circuit which was discussed briefly in Chapter 8. When the switch first contacts position 1, which will be called time t₁, the applied voltage causes the current to begin flowing in the direction indicated in Figure 10.36. Recall that if the inductance L is removed and only the resistor is present in the circuit, the current will instantaneously increase to a final value determined by Ohm’s law, the applied voltage divided by the resistance in the circuit. This current flow in a resistive circuit is shown in Figure 10.37. The current would rise immediately to 10 mA (10 volts divided by 1 kilohm). However, with an inductor in series with a resistor, the increasing current must pass through the inductor which opposes any change in current. What happens is that as the current initially attempts to increase from zero, the inductor sets up a counter-EMF. This counter EMF across the inductor inhibits the instantaneous rise of current as shown in Figure 10.38. The counter EMF of the inductor continues its opposition to the change of current until the current finally reaches its maximum value which is limited by the resistance in the circuit.

The rise of current in the inductive circuit follows an exponential curve as did the voltage rise across the plates of a capacitor. The maximum value of current is equal to the value of the applied voltage divided by the value of the resistance:

(10–7)

Thus, the maximum current in the example circuit of Figure 10.36 can be calculated:

Calculation of E_R

Since the voltage across the resistor is a direct result of the current flowing in the circuit, the voltage rise will appear similar to the rise of current. The voltage across the resistor will have a maximum value that can be expressed by Ohm’s law:

(10–8)

Thus the voltage across the resistor in the example dc RL circuit is calculated:

For a value of maximum current equal to 10 milliamperes, ER equals 10 volts.

Calculation of E_L

When the switch of the circuit in Figure 10.36 is placed in position 1, no current flows instantly, and no voltage appears across the resistor: I = 0; E_R = 0. Remember that this is a series circuit with a dc voltage applied. Because of this, Kirchhoff’s voltage law applies, and the applied voltage must equal the sum of the voltage drops in the circuit, which are the voltage across the resistor, E_R, plus the voltage across the inductor, E_L:

(10–9)

Because the voltage across the resistor is zero, the 10 volts applied to the circuit must all be dropped across the inductor. That is, the counter EMF is 10 volts and E_T = E_L. In addition, this is physically correct because the voltage across the inductor is equal to the inductance times the rate of change of current:

(10–10)

also is expressed in many instances as .

As a result, the voltage across the inductor, E_L, appears as shown in the waveform of Figure 10.39. Its highest value of 10 volts is at time t₁ where the rate of change of current is at its maximum.

As the voltage across the resistor increases, as shown in Figure 10.40, the voltage across the inductor must decrease to satisfy Kirchhoff’s law. When the current, I, has reached its maximum value, it stops changing value and the counter EMF across the inductor is zero. Recall that this is true because the counter EMF is a result of the inductor’s opposition to a change of current value. When the current reaches its maximum value the magnetic field surrounding the inductor is fully expanded as shown in Figure 10.41. The voltage across the resistor is now maximum, and it equals the applied voltage, thus obeying Kirchhoff’s voltage law.

Analysis of Circuit Current

Since the power supply has been disconnected, there is now no external voltage source to maintain current through the inductor and resistor. Thus, the magnetic field surrounding the inductor begins to collapse, and current begins to decrease in the circuit. As shown in Figure 10.43, the current direction is still the same, but it is becoming less and less in magnitude. The inductor, therefore, will generate a counter EMF to oppose the decrease in current. Thus, the collapsing magnetic field generates an EMF of the polarity shown in Figure 10.43 to keep current flowing in the circuit — to keep current from decreasing. The decay of current is exponential, as shown in Figure 10.44, because of the action of the inductor.

Analysis of E_R

The voltage across the resistor, which is a result of the current, decreases as the current decreases. This is evident in the decreasing curve after time t₂ in the waveform representing E_R in Figure 10.44. The counter EMF across the inductor has changed polarity in order to oppose the decrease in current.

Analysis of E_L

To help clarify this fact, remember that because the magnetic field around the inductor is collapsing, the flux lines are now cutting the windings of the inductor in the opposite direction from the direction that occurred when the magnetic field was increasing. When the magnetic field was increasing because of a changing current that was increasing, the counter EMF set up opposed the increase in current. Now, however, when the current decreases, the collapsing magnetic field sets up a counter EMF that now aids the current flow — to keep it going — to keep it from decreasing.Therefore, the counter EMF is of the opposite polarity because it aids the current flowing instead of opposing it.

Summary of Current and Voltage Changes as Current Decreases

In summary, then, when the switch is placed in position 2 in the circuit shown in Figure 10.42 the current exponentially decreases from 10 milliamperes to zero, as shown in Figure 10.45. The voltage across the resistor decreases from 10 volts to zero. And the voltage across the inductor changes polarity, and decreases in magnitude from −10 volts to zero.

Calculation of an RL Time Constant

In the example RL circuit with the switch in position 1, L = 10 millihenries, R = 1 kilohm. Thus, the time constant is calculated:

The time constant is 10 microseconds.

Five time constants are required for the current in the circuit to reach its maximum value. In the circuit the time constant is 10 microseconds. Therefore, five time constants equal 50 microseconds. Thus, it takes 50 microseconds for the current in the circuit to reach a value of 10 milliamperes. When the current decreases it also takes five time constants (50 microseconds) for the current to decrease to zero provided that the current flows through the same resistance and inductance. Figure 10.46 graphically illustrates the time required for the current to increase or decrease in the example RL circuit.

RL Circuit Summary

A typical RL circuit will be used to show you how to use the UTCC to determine voltages and currents in such a circuit. The RL circuit is shown in Figure 10.47. It is composed of an 8 henry inductor and 100 ohm resistor connected through a switch, S1, as shown, to an applied voltage of 20 volts.

The time constant is calculated:

The maximum voltage across the resistor and inductor is the applied voltage of 20 volts.

The maximum current is calculated using Ohm’s law:

The current will take five time constants to reach its final value. Since it takes 80 milliseconds for one time constant, it will take 400 milliseconds to reach its maximum value of 200 milliamperes, (5 × 80 ms = 400 ms).

Solving for Instantaneous Voltage and Current Values: Increasing Current

Suppose that one wants to know the value of the current, the voltage across the resistor, and the voltage across the inductor 160 milliseconds after the switch contacts position 1 as shown in Figure 10.48.

First, 160 milliseconds must be converted into terms of time constants. To do this, simply divide the time by the amount of time per time constant. In this case, the number of time constants represented by 160 milliseconds is 160 milliseconds divided by 80 milliseconds or two time constants.

Using the Universal Time Constant Chart the percent of maximum current is determined. This is about 86 percent. Therefore,

86 percent of 200 milliamperes is 172 milliamperes. Thus, the current in the circuit after 160 milliseconds is 172 milliamps.

The instantaneous voltage across the resistor can be calculated in the same way. However, since the voltage across the resistor is a result of the current in the circuit, the voltage across the resistor can be determined by using Ohm’s law and the value of the current.

In this example, E_R is 17.2 volts after 160 milliseconds.

The voltage across the inductor can be determined by using the chart. However, it can also be calculated. The calculation is based on the fact that the voltage across the resistor, E_R, plus the voltage across the inductor, E_L, must equal the applied voltage, E_T. Mathematically,

This expression can be rewritten to calculate E_L:

Therefore, if the voltage across the resistor is 17.2 volts:

The voltage across the inductor 160 milliseconds after t₁ is 2.8 V.

Solving for Instantaneous Voltage and Current Values: Decreasing Current

Once the current has reached maximum, the switch in the circuit, as shown in Figure 10.49, is changed to position 2. For an example, the voltages and currents in that circuit will be calculated for a time 80 milliseconds after the switch has contacted position 2. The waveforms for E_R, E_L, and I are shown in Figure 10.50.

First, the number of time constants after t₂ is calculated.

Using the Universal Time Constant Chart, the percent to which the maximum current has decayed after one time current is determined as 37 percent. As shown in Figure 10.51, the current value after 80 milliseconds (one time constant) can be calculated:

Thus, the current 80 milliseconds after time t₂ is 74 milliamperes. This is shown graphically in Figure 10.52.

The voltage across the resistor, E_R after 80 milliseconds is equal to 37 percent of E_Rmaxand can be calculated:

This is shown graphically in Figure 10.53. Remember that Kirchhoff’s voltage law must be satisfied; therefore, the voltage across the inductor, EL, after 80 milliseconds is 37 percent of E_Lmax, but negative.

This is shown graphically in Figure 10.54.

Consequently, in the example circuit 80 milliseconds after t₂, current will be 74 mA, the voltage across the resistor will be 7.4 volts and the voltage across the inductor will be −7.4 volts

a. R = 60 k, C = 0.02 μF, Iτ = ________

b. R = 5 k, C = 30 μF, Iτ = ________

a. 

b. 1τ = R × C = 5 × 10³ × 30 × 10⁻⁶ =0.15 s

a. 1τ = 5 ms, R = 30 k Ω, C = ________

b. 1τ = 30 μs, R = 47 k Ω, C = ______

a. 

b. 

a. 1τ = 5 s, C = 0.1 μF, R = ______

b. 1τ = 5 ms, C = 0.1 μF, R = ________

a. 

b. 

a. I_max=_____

b. E_Cmax = ______

c. E_Rmax = ______

d. 5τ = ______

a. 

b. E_Cmax = E_A = 20 V

c. E_Rmax = E_A = 20 V

d. 5τ = 5 × 1τ = 5 × 5 ms =25 ms

a. time = 6 ms

b. time = 30 ms

c. time = 50 ms

d. time = 80 ms

a. 

b. 

c. 

d. 

a. number of time constants equals 0.7τ

b. number of time constants equals 1.3τ

c. number of time constants equals 2.5τ

d. number of time constants equals 3.8τ

a. time = 0.7 × 50 μs =35 μs

b. time = 1.3 × 50 μs =65 μs

c. time = 2.5 × 50 μs =125 μs

d. time = 3.8 × 50 μs =190 μs

a. E_c = ______

b. E_R = ______

c. I = ______

(1) Calculate the number of time constant represented by 12 ms:

(2) Use UTCC to determine percent of maximum circuit value after 2.4τ:

a. E_C = 0.9 × 20 V =18 V

b. E_R = 0.1 × 20 V =2 V

c. I = 0.1 × 4 ma =0.4 mA

a. I_max ________

b. E_Lmax = ________

c. E_Rmax ________

d. 5τ = ________

a. time = 3 ms

b. time = 10 ms

c. time = 20 ms

d. time = 38 ms

a. E_L = ________

b. E_R =________

c. I = __________

a. I_max ______

b. E_Cmax = ______

c. E_Rmax = ______

d. 5τ = ______

a. 1.5τ

b. 3τ

c. 4.0τ

d. 4.5τ

a. 0.8 ms

b. 1.2 ms

c. 3.2 ms

d. 5.0 ms

a. E_C = ______

b. E_R = ______

c. I = ______