Resistive Circuits
In previous chapters, ac was discussed in general terms. With this chapter, a more detailed investigation of actual circuits with an ac voltage applied will begin. This chapter describes how to determine voltages and current in series, parallel, and series-parallel resistive circuits when an ac voltage is applied. Peak, peak-to-peak, and rms voltage, current and power values are calculated. The phase and frequency of circuit current and component voltages in ac resistive circuits are analyzed.
At the end of this chapter, you should be able to:
1. Analyze the phase and frequency relationships of current and voltage in an ac resistive circuit.
2. Explain the relationship of peak, peak-to-peak, and rms values for an ac resistive circuit and be able to convert from one value to another.
3. Calculate the instantaneous voltage and current values in an ac resistive circuit and plot the results.
4. Analyze series, parallel, and series-parallel ac circuits as shown and calculate voltage, current, and power values.
REVIEW OF A FEW BASICS
Recall that when a fixed dc voltage is applied to a resistor as shown in Figure 5.1, current flows in only one direction and its value is determined by the value of the voltage and the resistor in the circuit. According to Ohm’s law, the current is equal to the voltage divided by the resistance.
Therefore, the current in the circuit of Figure 5.1 is 10 milliamperes.
If the connections to the battery are reversed as shown in Figure 5.2, the direction of the current reverses, but the current value is the same. The current is still 10 milliamperes.
Figure 5.2 The Resistive DC Circuit of Figure 5.1 with Battery Connections Reversed
If, now, a sinusoidal voltage is applied to the same circuit with the resistor as shown in Figure 5.3, the voltage in the circuit will be changing constantly. This is shown in Figure 5.4. Ohm’s law still applies and can be used at each instant of time to calculate the current just as it was used to calculate the current when a constant dc voltage was applied. This is an important point of this lesson and bears repeating. When a known ac voltage is applied to a purely resistive circuit Ohm’s law can be used at any instant of time to calculate the current at that same instant of time.
Calculating Instantaneous Voltages and Currents
Recall that the instantaneous value of the voltage for a sinusoidal waveform can be calculated using this equation:
The instantaneous voltage is equal to the peak voltage times the sine theta.
Figure 5.5 shows a sinusoidal waveform that can be described with equation 5–2. The peak amplitude of that waveform is 10 volts. Since the sine of zero degrees is zero, at zero degrees the voltage is zero. At 45 degrees the voltage has risen to 7.07 volts calculated as follows:
At 90 degrees the voltage has risen to 10 volts. The waveform voltage declines as the angle theta increases. At 135 degrees the voltage is once again 7.07 volts, and at 180 degrees, the voltage is zero.
During the negative alternation, the voltage amplitude varies in the same manner as for the positive alternation, but with opposite — negative — polarities.
Ohm’s law can be used with each of these instantaneous voltage values to calculate the instantaneous current value just as if each instantaneous voltage were produced by a battery of voltage Eiplaced in the circuit at the right moment as shown in Figure 5.6. At zero degrees (point A of Figure 5.5) the instantaneous value of the voltage is zero. Therefore, according to Ohm’s law, the instantaneous current at zero degrees equals the voltage divided by the resistance:
At 45 degrees (point B of Figure 5.5) the instantaneous voltage, ei, is 7.07 volts. This can be represented by a 7.07 volt battery as shown in the circuit in Figure 5.7. Using Ohm’s law as before, the current at 45 degrees is:
The current flowing at that instant in the cycle is 7.07 milliamperes.
Instantaneous current values continue to vary this way throughout the entire cycle as shown in Figure 5.8. In the negative alternation, however, the current is plotted as negative because its direction of flow is opposite to the direction called positive. If all of these plotted points are connected with a smooth curve, as shown in Figure 5.9, it is found that the current is also sinusoidal.
Relationship of Peak, Peak-to-Peak, and RMS Values
Recall that the amplitude of a sinusoidal voltage can be specified in three ways: peak, peak-to-peak, and rms. The voltage waveform in the previous example has a peak amplitude of 10 volts as shown in Figure 5.10. Thus, it has a peak-to-peak amplitude of 20 volts.
It also has an rms amplitude of
Thus, its rms voltage is 7.07 volts.
Because the current is also sinusoidal, it too has peak, peak-to-peak, and rms amplitudes. In the example, the current has a peak amplitude of 10 milliamperes as shown in Figure 5.11. It, therefore, has a peak-to-peak amplitude of 20 milliamperes (twice the peak amplitude); and it has an rms amplitude of 7.07 milliamperes (0.707 times the peak amplitude).
There should be no doubt that Ohm’s law applies to an ac resistive circuit. However, with ac circuits you must be aware of the way in which the circuit values are specified. For example, if the peak voltage of the waveform is divided by the resistance, the result is peak current. In the example the peak current is 10 milliamperes calculated as follows:
Peak-to-peak voltage divided by the resistance yields peak-to-peak current. In the example this is 20 milliamperes calculated as follows:
And the rms voltage divided by the resistance yields rms current. In the example this is 7.07 milliamperes.
Converting Values
An applied voltage or current can be specified in one of these three ways. But suppose a current or voltage value is specified as peak and its rms amplitude is needed. As you may recall from previous discussions, it is easy to convert the peak specification to its rms value by calculation. The rms voltage or current is equal to 0.707 times the peak voltage or peak current.
Or it may be necessary to convert from any one of these peak, peak-to-peak, and rms specifications to another. A table, Table 5.1, of conversion equations is included to enable you to convert from any one of these specifications to another for either voltage or current.
Voltage and Current Phase Relationship
If the voltage waveform and the current waveform are examined together as shown in Figure 5.12, it can be seen that when the voltage is at a maximum or peak, the current is also at a maximum or peak. The current flow is less as the value of the instantaneous voltage decreases. When the voltage is at zero, the current is at zero. Also note that the current is flowing in a direction determined by the polarity of the voltage. That is, in the example shown, a positive 10 milliamperes flows when the voltage is a positive 10 volts and a negative 10 milliamperes flows when the voltage is a negative 10 volts.
As shown in Figure 5.13, the voltage and current peak at the same time. Because they both peak at the same time, and are zero at the same time, and move in the same direction, the voltage and current are in phase. This means there is no phase difference between them. Their amplitudes and polarities are the same for each instant of time. Therefore, the phase angle between them is zero.
As you recall from previous chapters, this phase relationship can be represented with a pair of phasors as shown in Figure 5.14. One phasor represents the phase of the voltage; the other phasor represents the phase of the current. One phasor is drawn on top of the other to show that there is no phase angle or phase difference. The vector lengths have no real meaning because they have different units. To compare two vector lengths meaningfully the units of both vectors must be the same — both must represent volts, amperes, ohms, milliamperes, etc. In Figure 5.14 one vector represents volts while the other represents amperes. A comparison of the amplitude of the two quantities means nothing; only the phase relationships of the two quantities is of importance. The phase relationship of currents and voltages is very significant in ac circuits. It will be discussed in detail later in the chapter.
There’s another important point concerning the two waveforms. Since one cycle of the voltage waveform corresponds to one cycle of the current waveform as shown in Figure 5.13 the frequency of both is the same. Therefore, no matter what frequency of voltage is applied to any circuit of this type, all voltages and currents in that circuit will have the same frequency.
CIRCUIT ANALYSIS
As you know, resistive circuits can consist of several resistors in series, as shown in Figure 5.15, in parallel as shown in Figure 5.16, and in series-parallel combinations as shown in Figure 5.17.
AC resistive circuits can be analyzed just like dc resistive circuits, and the circuit voltage and current values determined in the same manner. This is possible because currents and voltages are in phase in this type of circuit.
The voltage or current values can be described in peak, peak-to-peak, or rms values. You must make the necessary conversions to obtain your answers in the desired units.
Series Resistive AC Circuit
The first circuit to be analyzed is the series circuit shown in Figure 5.15. There are two resistors in this circuit in series with an applied ac voltage of 10 volts peak. To determine the voltages and currents in this circuit, first the total resistance must be calculated. The total resistance in the circuit is the sum of the individual resistances,
The total resistance of the circuit is 10 kilohms. Once RThas been determined the voltage and currents in the circuit can be calculated.
Since the total resistance we calculated is 10 kilohms, to find the total current in the circuit, divide the applied voltage by the total resistance:
The total current in the circuit is 1 milliampere peak. Since this is a series circuit, this current is the only current in the circuit. It follows only one path — through both resistors in the circuit.
Next, the voltage drops across each resistor can be found using Ohm’s law. The voltage drop across R1 is equal to the current through R1, which is the total current, times the value of R1:
The drop across R2 is calculated similarly:
Notice that Kirchoff’s voltage law is satisfied because the voltage drop across R1, 2 volts peak, plus the voltage drop across R2, 8 volts peak, equals the applied voltage, 10 volts peak:
The circuit source is an ac voltage and this voltage is specified in terms of its peak amplitude. It is 10 volts peak. The peak value of voltage was used to calculate the voltage drops and current in the circuit. Therefore, the currents and voltage drops calculated are peak currents and peak voltages. For this reason the written calculated values include the word peak.
If the applied voltage had been specified as 10 volts peak-to-peak, then the circuit currents and voltage drops would also be peak-to-peak values because calculations would be performed using the 10 volts peak-to-peak. The specification peak-to-peak would be written along with the calculated value as shown in Figure 5.18. Similarly, if the applied voltage had been specified as rms, then the voltage drops and currents calculated using this rms value would also be rms values.
Figure 5.18 Specification of EAas Peak, Peak-to-Peak, or RMS, Determines Specification of Other Circuit Values
In summary then, to determine the voltages and currents in a resistive circuit with an ac voltage source, calculate the voltages and currents in the circuit the same as these factors are calculated in a dc series circuit. The circuit voltages and currents are specified the same as the ac applied voltage: peak, peak-to-peak, or rms.
Once all voltages and currents in a circuit have been calculated using one amplitude specification, the other two types of amplitude values of these same circuit currents and voltages can be determined readily by directly converting each value using the appropriate conversion equation.
Using the series circuit problem we just solved as an example, a table of values can be created as shown in Figure 5.19. In the column labeled peak are all the peak values calculated for the circuit. The peak-to-peak value is twice their peak value. The rms values are 0.707 of the peak values. For example,
the rms values have been calculated using equations 5–3 and 5–4 as follows:
Erms = 0.707 Epk
EArms = (0.707 × 10 V) = 7.07 V
ER1rms =(0.707)(2 V) = 1.414 V
ER2rms = (0.707)(8 V) = 5.656 V
Irms = 0.707 Ipk
ITrms = (0.707)(1 mA) = 0.707 mA = 707 μA
In other words, in the circuit, the applied voltage is 10 volts peak, which equals 20 volts peak-to-peak or 7.07 volts rms. These values are shown graphically in Figure 5.20.
The 2 volt peak voltage drop across R1 equals 4 volts peak-to-peak or 1.414 volts rms as shown in Figure 5.21. The three values, 8 volts peak, 16 volts peak-to-peak, and 5.656 volts rms are shown in Figure 5.22 for ER2.
The circuit current shown graphically in Figure 5.23, is 1 milliampere peak. This equals 2 milliamperes peak-to-peak or 707 microamperes rms.
Remember that all voltages and currents in this purely resistive circuit are in phase and of the same frequency. Also note in Figure 5.19 that in each column the value of ER1and ER2add up to the applied voltage in each column.
If a circuit has the source voltage specified in a peak-to-peak value, but answers are required in rms values, the voltages and currents can be determined one of two ways: either by using the peak-to-peak values and then converting all answers to rms values or by converting the source voltage to an rms value before performing any other calculations. Then all of the following calculations will be in rms values.
Also an important point to remember is that whenever the designation VAC is used, it is understood, by convention, to mean an rms value.
Parallel Resistive AC Circuit
Now that you have been shown how to determine all values of voltage and current in a series resistive circuit, the next step is to show you how this is done with a parallel resistive circuit.
In the resistive circuit of Figure 5.16 and repeated as Figure 5.24, 30 volts peak-to-peak ac voltage is applied to a 3-kilohm and 6-kilohm resistor connected in parallel. First the circuit will be analyzed in terms of peak-to-peak amplitudes, and then all voltages and currents will be converted to their peak and rms amplitudes.
In the resistive circuit of Figure 5.16 and repeated as Figure 5.24, 30 volts peak-to-peak ac voltage is applied to a 3-kilohm and 6-kilohm resistor connected in parallel. First the circuit will be analyzed in terms of peak-to-peak amplitudes, and then all voltages and currents will be converted to their peak and rms amplitudes.
Just like dc parallel circuits, in ac parallel circuits the applied voltage (in this example 30 volts peak-to-peak) is across each resistor. If this applied voltage is divided by each branch resistor’s value, the peak-to-peak branch current for each branch can be determined.
The peak-to-peak current through resistor R1 is 10 milliamperes.
The peak-to-peak current through resistor R2 is 5 milliamperes.
The total current is the sum of the branch currents. Since the branch currents are known, they can be added to determine the total current.
The total current is 15 milliamperes peak-to-peak.
Therefore, as in a dc parallel circuit, the current through each branch of an ac parallel circuit may be different depending on the resistor values, while the voltage drop across each resistor is the same. Also, the algebraic sum of the branch currents equals the total current.
To verify that this is the correct total current the applied voltage can be divided by the total resistance in the circuit. Recall that the total resistance for two parallel resistors is their product divided by their sum. A special symbol is used to indicate the parallel connection. Therefore,
indicates that R1is in parallel with R2. Thus,
is the expression to calculate the total resistance of two resistors connected in parallel.
In the example circuit of Figure 5.24, the total resistance, calculated by using equation 5–5, is:
The total current in the circuit is calculated using Ohm’s law:
This calculation shows that 15 milliamperes peak-to-peak current is the total current and verifies previous calculations. Using these peak-to-peak values for the circuit, the peak and rms values can be calculated. Peak values are one-half the peak-to-peak values. The rms values are 0.707 times the peak value or 0.3535 times the peak-to-peak value. The calculations for the peak values are as follows:
The calculations for the rms values are as follows:
EArms= (0.3535)(30 Vpp) = 10.6 V
IR1rms= (0.3535)(10 mApp) = 3.54 mA(3.535)
IR2rms= (0.3535)(5 mApp) = 1.77 mA(1.7675)
ITrms= (0.3535)(15 mApp) = 5.30 mA
Once calculated they can be written in tabular form as shown in Figure 5.25. Remember that all voltages and currents are in phase, and that they all have the same frequency. Also, the branch currents sum to the total current in each column.
Figure 5.25 Calculated Values of the Circuit of Figure 5.24
Series-Parallel Resistive AC Circuit
Now that a series resistive circuit and a parallel-resistive circuit have been analyzed, the next circuit to be studied is a series-parallel resistive circuit shown in Figure 5.26. This is the same circuit as Figure 5.17.
In the circuit, 24 volts ac is applied to a series-parallel combination of four resistors. Recall that if an ac voltage is specified as VAC then the value is an rms amplitude. That is, the applied voltage is 24 volts rms.
Simplified Circuit
The series-parallel combination of resistors consists of R1and R4in series with the parallel combination of R2and R3. To begin determining the voltages and currents in this circuit, as with dc circuits, first the circuit is simplified by finding the equivalent resistance of R2 and R3 sin parallel. It is calculated using the form of equation 5–5.
Thus, the parallel combination of R2and R3(R23) can be thought of as a 3 kilohm resistor substituted in the circuit as shown in Figure 5.27. As you probably realize, the circuit is now a simple series circuit. Therefore, its total resistance is equal to the sum of the individual resistances.
Total Resistance
The total resistance equals 2 kilohms for R1, plus 3 kilohms for the parallel combination of R2 and R3, plus 3 kilohms for R4:
Total Current
To determine the total current supplied by the ac voltage generator to the circuit the applied voltage is divided by the total resistance. 24 volts divided by 8 kilohms is:
The total current is 3 mA rms. This total current flows through R1, through the parallel combination of R2 and R3, and through R4. Therefore, IR1 and IR4 are 3 mArms.
Voltage Drops
Ohm’s law is now used to determine the voltage drop across each resistance. It is the value of the resistance multiplied by the current through it. Thus,
The voltage drop across R1 is 6 volts rms.
The voltage drop across R2 in parallel with R3is equal to the total current times the value of the equivalent parallel resistance R23:
The voltage drop across R23 is 9 volts rms.
The voltage drop across R4 is calculated in a similar manner as:
Branch Currents
Now the current through R2 and R3 can be calculated. Since there are 9 volts across R23, the parallel combination of R2 and R3, there are 9 volts across R2 and 9 volts across R3. Thus, you see that this parallel combination is treated just like the separate parallel circuit. Using Ohm’s law the value of the current through R2 and through R3 (these are like the branch circuits in the separate parallel circuit) can be determined by dividing the voltage across either resistor by the value of the resistor.
For R2, the value of the current is:
IR2 of Figure 5.26 is 2.25 milliamperes rms.
The current through R3 is calculated similarly:
IR3 of Figure 5.26 is 0.75 milliamperes rms.
Voltage Chart
Figure 5.28 shows a tabulation of the voltages in this series-parallel circuit. These values are in the rms column since the applied voltages and all the voltage drops calculated are rms. These rms values can be converted directly to both peak, and peak-to-peak values. The peak values are simply 1.414 times the rms values. To determine the peak-to-peak amplitude, either the peak values can be doubled, or the rms values can be multiplied by 2.828. Note that in each column the voltage across R2 and R3 is the same since they are in parallel, and this voltage along with the other series circuit voltages ER1 and ER4, add up to the applied voltage.
Figure 5.28 Calculated Voltage Values for the Circuit of Figure 5.26
Current Chart
Figure 5.29 shows a tabulation of the values of the currents for the circuit for Figure 5.26. The rms current values are the original ones calculated and these values are used to convert to peak and peak-to-peak current values. The peak current calculations are as follows:
Figure 5.29 Calculated Current Values for the Circuit of Figure 5.26
IR2pk =(1.414)(2.25 mA) =3.18 mA
IR3pk =(1.414)(0.75 mA) =1.06 mA
The peak-to-peak current values are as follows:
The results are a bit different through the voltage chart. Now the two branch currents through the parallel combination of R2 and R3 add to the total series current, which is the same through R1 and R4.
The effects of ac voltages and currents in series, parallel, and series-parallel resistive circuits have now been discussed thoroughly. However, there is another important consideration in these circuits — the power dissipated.
POWER DISSIPATION
In ac circuits there are three kinds of power:real, reactive, and apparent power. The kind of power that exists in purely resistive circuits is what is called real power.Real power is power that is dissipated in the form of heat and is measured in watts as illustrated in Figure 5.30.
Recall, as shown in Figure 5.31, that in dc circuits the amount of power dissipated in watts can be calculated by multiplying the voltage across a resistor by the current passing through it.
The power, P, is in watts, when the voltage, E, is in volts, and the current, I, is in amperes. It turns out that equation 5–6 is true whether a dc voltage or an ac voltage is applied. This is shown in Figures 5.30 and 5.31.
There is one difference, however. The power in an ac resistive circuit can be specified as peak, peak-to-peak, or rms power. If the peak voltage across a resistor is multiplied by the peak current passing through it, the result is peak power:
If the rms voltage across a resistor is multiplied by the rms current through it, the result is the rms power value:
If the peak-to-peak voltage across a resistor is multiplied by the peak-to-peak current through it, the result is the peak-to-peak power value:
Each of these three different power values is of different importance in ac circuit considerations. The peak-to-peak power dissipation has little significance because power does not depend on the polarity of a voltage or the direction of the current. The peak power dissipation also is usually only significant in certain applications because peak power is an instantaneous value occurring only at the instant the voltage and current are at their peak values. The most significant power specification in an ac circuit is the rms power. It is more significant because it is the type of ac power that can be equated with dc power. (Remember that rms was defined to provide the same equivalent heating effect as dc.)
Using equation 5–6 the power dissipated by a resistor is determined by multiplying the voltage across the resistor by the current flowing through it.
Thus, this equation can be used to calculate the rms power dissipated in series, parallel, and series-parallel ac resistive circuits. For example, it can be used to calculate the power dissipated in each of the four resistors in our previous series-parallel circuit shown again in Figure 5.32. This is the same circuit as Figure 5.26 with the rms current calculated values added. Keep in mind that it is a series-parallel circuit and that the principles discussed can be applied to determine the rms power dissipated in simple series or simple parallel circuits as well.
The voltage drop across R1 of 6 volts rms was previously calculated. It is used along with the current through R1 to calculate the power dissipated in R1 as follows:
The rms power dissipated by R1 is 18 milliwatts.
The power being dissipated by resistors R2, R3, and R4 can be calculated similarly. Recall that the voltage drop across the parallel combination of R2 and R3 was calculated as 9 volts and across R4 as 9 volts. Therefore,
Therefore, the rms power dissipated by each of these resistors is: R2 = 20.25 milliwatts; R3 = 6.75 milliwatts; and R4 = 27 milliwatts.
The total power being dissipated within the entire circuit equals the total applied voltage times the total current.
The total is 72 milliwatts which verifies the previous calculation for total power.
All individual power dissipation values should add up to the total power calculated:
The total is 72 milliwatts which verifies the previous calculation for total power.
Remember that in all of these power calculation examples, calculations have been rms power.
SUMMARY
In this chapter, purely resistive circuits with an ac voltage applied were analyzed. Calculations were performed to show you how to determine the amplitudes of the voltages, currents, and power dissipations in series, parallel, and series-parallel ac resistive circuits. You were also shown how to convert between the peak, peak-to-peak, and rms amplitude specifications of these quantities.
It should be remembered that all voltages and currents in these circuits have the same period and therefore, the same frequency. In pure resistive ac circuits there is no phase difference between the voltages and currents, so the voltage and current are said to be “in phase.”
1. Given the peak-to-peak circuit values shown in the table, complete the table by converting to peak and rms values.
a. ER1pk = 0.5 × ER1pp = 0.5 × 50 V =25 v
b. ER2pk = 0.5 × ER2pp = 0.5 × 30 =15 V
c. ER3pk = 0.5 × ER3pp = 0.5 × 27 V =13.5 V
d. ER4pk = 0.5 × ER4pp = 0.5 × 14 V =7 V
e. ER1rms = 0.707 × ER1PK = 0.707 × 25 V =17.7 V
f. ER2rms = 0.707 × ER2pk = 0.707 × 15 V =10.6 V
2. For the circuit below, calculate the instantaneous voltage and current values specified by the table:
| Degrees | ei | ii |
| 0 | a. ______ | h. ______ |
| 30 | b. ______ | i. ______ |
| 60 | c. ______ | j. ______ |
| 90 | d. ______ | k. ______ |
| 120 | e. ______ | l. ______ |
| 150 | f. ______ | m. ______ |
| 180 | g. ______ | n. ______ |
Epk = 0.5 × Epp = 0.5 × 15 Vpp =7.5 V
Then the instantaneous voltage and current values can be calculated.
a. ei(0°) = sin 0° × Epk = 0.00 × 7.5 V =0 V
b. ei(30°) = sin 30° × Epk = 0.5 × 7.5 V =3.75 V
c. ei(60°) = sin 60° × Epk = 0.866 × 7.5 V =6.5 V
3. For the circuit in Example 2, plot the current and voltage waveforms based on the calculated instantaneous values.
4. For the following circuit and the typical circuit values shown, determine the circuit values specified in rms values.
a. RT = _______________________
b. IT = _______________________
c. IR1= _______________________
d. IR2= _______________________
e. IR3= _______________________
f. ER1 = _______________________
g. ER2 = _______________________
h. ER3= _______________________
i. PR1 = _______________________
j. PR2 = _______________________
a. RT =R1 +R2 +R3 =3 kΩ+4.5 Ω+1.7 kΩ=9.2 kW
f. ER1=IR1 ×R1 =4.35 mArms ×3 kΩ =13.0 VAC
g. ER2 =IR2 ×R2 =4.35 mArms ×4.5 kΩ =19.6 VAC
h. ER3 =IR3 ×R3 =4.35 mArms ×1.7 kΩ=7.4 VAC
i. PR1 =IR1 ×ER1 =4.35 mArms ×13 VAC =56.5 mWrms
j. PR2 =IR2 ×ER2 =4.35 mArms ×19.6 VAC =85.2 mWrms
5. For the following circuit and typical circuit values shown, determine the circuit values specified in peak values.
a. IR1= _______________________
b. IR2= _______________________
c. IR3= _______________________
d. IT= _______________________
e. RT= _______________________
f. ER1= _______________________
g. ER2= _______________________
h. ER3= _______________________
i. PR1= IR1 × ER1= 6.67 mApk × 20 Vpk =133 mWPk
j. PR2= IR2 × ER2= 4.44 mApk × 20 Vpk =88.8 mWPk
6. For the following circuit and typical circuit values shown, determine the circuit values specified in peak-to-peak values.
c. IR1 = _____________________
d. IR2 = _____________________
e. IR3 = _____________________
f. ERI = _____________________
g. ER2 = _____________________
h. ER3 = _____________________
i. PR1 = _____________________
j. PR2 = _____________________
1. Given the rms circuit values below, complete the table by converting to peak and peak-to-peak values.
2. For the following circuit, calculate the instantaneous voltage and current values specified by the table:
| degrees | ei | ii |
| 180 | a. ______ | h. ______ |
| 210 | b. ______ | i. ______ |
| 240 | c. ______ | j. ______ |
| 270 | d. ______ | k. ______ |
| 300 | e. ______ | l. ______ |
| 330 | f. ______ | m. ______ |
| 360 | g. ______ | n. ______ |
3. For the circuit in Problem 2, plot the current and voltage waveforms based on the calculated instantaneous values.
4. For the following circuit and typical circuit values shown, determine the circuit values specified in peak-to-peak values.
a. RT = ______________________
b. IT = ______________________
c. IR1 = ______________________
d. IR2 = ______________________
e. IR3 = ______________________
f. ER1 = ______________________
g. ER2 = ______________________
h. ER3 = ______________________
i. PR1 = ______________________
j. PR2 = ______________________
5. For the following circuit and typical circuit values shown, determine the circuit values specified in rms values.
6. For the following circuit and typical circuit values shown, determine the circuit values specified in peak values.
1. Given the peak circuit values shown in the table, complete the table by converting to peak-to-peak and rms values.
2. For the following circuit, calculate the instantaneous voltage and current values specified by the table.
| degrees | ei | ii |
| 0 | a. _______ | j. _______ |
| 45 | b. _______ | k. _______ |
| 90 | c. _______ | l. _______ |
| 135 | d. _______ | m. _______ |
| 180 | e. _______ | n. _______ |
| 225 | f. _______ | o. _______ |
| 270 | g. _______ | p. _______ |
| 315 | h. _______ | q. _______ |
| 370 | i. _______ | r. _______ |
3. For the circuit in problem 2, plot the current and voltage waveforms based on the calculated instantaneous values.
4. For the following circuit and the typical circuit values, determine the circuit values specified in peak values.
5. For the following circuit and typical circuit values shown, determine the circuit values specified in peak-to-peak values.
6. For the following circuit and typical values shown, determine the circuit values specified in rms values.
a. RT = ______________________
b. IT = ______________________
c. IR1 = ______________________
d. IR2 = ______________________
e. IR3 = ______________________
f. ER1 = ______________________
g. ER2 = ______________________
h. ER3 = ______________________
i. PR1 = ______________________
j. PR2 = ______________________