ALTERNATING-CURRENT SERIES CIRCUIT
OBJECT: To study the electrical characteristics of an AC circuit containing a resistor, and inductor, and a capacitor in series.
METHOD: A resistor, and inductor and a capacitor are connected in series and joined to a source of alternating current. Measurements are made of the current in the circuit and the voltages across the elements in the circuit. Phase diagrams are constructed and analyzed for several values of current.
THEORY: In a direct-current (DC) circuit the electrons in a metallic conductor flow continuously in the same direction. In an alternating-current (AC) circuit they flow in one direction for only a short time, then reverse and move in the opposite direction for an equally short time before making another reversal. These reversals usually occur many times per second. In a 60 cycle circuit there are 120 reversals per second. The graph of current against time for one cycle or period T is the typical sine curve of Fig. 1. The plot of voltage against time is a similar curve.
Figure 1: Sine curve for one cycle of an alternating current.
Instantaneous and Effective Values of Current and Voltage:
The instantaneous values of the voltage v and current i in an AC circuit are continuously changing. The manner in which each varies with time is shown in the equations:
I = Io sin (2f t) (1)
V = Vo sin (2f t) (2)
where Io and Vo are the maximum values of current and voltage, t is time and f is the frequency in cycles per second. Meters placed in the circuit are usually designed to read the effective values of current and voltage. The effective value of an AC circuit current is defined as that value of a steady current which will develop heat at the same rate in the same resistor. The effective value of current and voltage (called rms values), designated as Irms and Vrms are 0.707 of their respective maximum values. Thus, a 120-volt, 60 cycle AC circuit fluctuates between approximately + 170 volts and - 170 volts in each cycle.
Phase Relationships: When a pure resistance carries an alternating current, the resulting potential difference or voltage across its terminals is in phase with the current. This means that both are zero at the same instant and both arrive at their maximum value at the same instant, Fig. 2a. The phase diagram, Fig 2b, records the potential drop of the resistor in the same direction (phase) as the current. Both are usually drawn horizontally.
Figure 2: Resistance drop in phase with current.
In a pure inductance, the applied voltage on the terminals leads the current by 90. The phase diagram, therefore, records voltage across an inductor as a line directed vertically upward. The terminal voltage across a capacitance lags the current by 90 and is represented in a phase diagram by a line drawn vertically downward. Figure 3b shows the phase diagrams for the voltages in the resistor, inductor and capacitor. The symbols in the phase diagram correspond to the observed voltage readings indicated in the circuit diagram of Fig. 3a.
Figure 3: Phase relationship of voltage in AC circuit containing R, L, and C in series.
The angular relationships of these potential differences indicate that the voltages observed in a series AC circuit, in which the current is the same in all parts, may be treated as a vector sum. Thus, the voltage V, across all three elements of the circuit in Fig. 3a, is the vector sum of VR, VL, and VC (the potential drops across the resistor, inductor, and capacitor respectively). This is shown in the phase diagram Fig. 3c. Note that the potential difference measured across an element in the circuit may be greater than the voltage across the entire circuit.
Effect of Inductance in a Circuit: The 90 angle of lag of current behind voltage in an inductor results from the changing magnetic field about the coil. This changing magnetic field produces an induced emf i expressed by:
i = -L di/dt (3)
where L is the inductance and di/dt is the time rate of current change. The induced emf is in volts when L is in henries and di/dt is in amperes per second. The value of di/dt for an AC current may be obtained by differentiating Eq. (1). Substituting this value for di/dt in Eq.(3) gives:
i = -2f L Io cos (2f t) (4)
The impressed voltage to establish the current must be equal in magnitude and opposite in direction to the induced emf, namely,
V = 2f L Io cos (2f t) (5)
This equation compared with the current equation, Eq.(1), shows a phase difference of 90°, with the voltage leading the current.
Effect of Capacitane in a Circuit: A capacitor connected into a DC circuit provides a momentary current only while the plates are charging to the potential of the power source. When this is achieved, no additional charge flows. In an AC circuit the plates are charged alternately in opposite directions and thus there is a continuing back and forth surge of charge (current) in the circuit. The current at any instant is given by
I = dq/dt (6)
where dq is the charge transferred in time dt. Since the capacitance C of a capacitor is C = q / V,
I = C dV/dt (7)
The value of dV/dt for an AC circuit may be obtained by differentiating Eq. (2). Inserting this value of dV/dt in the above equation gives:
I = 2f C Vo cos (2f t) (8)
I = 2f C Vo sin (2f t + 90°) (9)
Comparison of Eq.(9) with Eq.(2) shows that there is again a 90° phase angle with current leading the applied voltage or the voltage lagging the current. This appears reasonable recognizing that after the voltage passes through its maximum value the charged capacitor plates will begin to discharge.
Reactance: The maximum values of current and voltage in Eqs. (5) and (9) occur when cos (2f t) = 1. Since the maximum values of current and voltage are proportional to their effective values,
VL / I = 2f L = XL (10)
VC / L = 1 / 2f C = XC (11)
The reactance of an inductor or capacitor is defined as the ratio of the voltage across its terminals to the current produced. Thus, XL is the inductive reactance of an inductor and XC is the capacitive reactance of a capacitor. The reactance of each is expressed in ohms when L is in henries, C in farads and f in cycles per second.
Note that the voltage across an inductor, called reactance drop, is given by:
VL = I XL (12)
VC = I XC (13)
The voltage drop across a resistor, resistance drop, is given by Ohm's law:
VR = I R (14)
The resistance drop and the reactance drop are always at right angles to each other.
Impedance: The current in a DC circuit, for a given power source, is determined by the resistance R of the circuit. Its value is expressed by I = V/R. In an AC circuit the current is determined not only by the resistance but also by the reactance which may be present. Thus, R is replaced by Z, the impedance of the circuit, giving the expression Z = V / I. A graph, plotting terminal voltages against currents in a circuit, provides a curve whose slope is the impedance of the circuit.
Figure 4: Impedance diagram of circuit Fig. 3a
Replacing the potential drops of Fig. 3c by their corresponding values, namely V = I Z, VR = I R, VL = I XL and VC = I XC, and observing that the current values I are the same in all parts of a series circuit, gives the impedance diagram Fig 4. It is evident from this figure that the impedance of the circuit is expressed by
Z = R2 + (XL - XC)2 (15)
The current in the circuit Fig. 3 is, therefore,
I = V / Z (16)
Note that when the circuit contains neither inductive nor capacitive reactance, the expression reduces to the simple form I = V / R. When XL = XC, the current is a maximum and also equal to V / R. This condition is known as resonance!
Phase Angle and Power Factor: In general, an impedance consists of a combination of a resistance and a reactance. Thus, the impedance has an angle , Fig. 4, called the phase angle, associated with it. This is the angular relationship between V and I in the circuit or portion of the circuit considered. The phase angle may be either a lag or a lead angle depending on the relative values of XL and XC. Power is consumed only by the resistive component of the impedance. The phase angle between the current and voltage, therefore, enters the power equation.
P = I V cos (17)
The term cos is called the power factor of the circuit. When = 0 , the value of cos = 1. This is the power expression for a DC circuit. When = 90 , the value of cos = 0 and the power dissipated in the circuit, regardless of the magnitude of I or V, is zero.
Laboratory Phase Diagrams: The ideal case of pure resistance, inductance and capacitance, so far considered, is never attained in the laboratory. A good quality capacitor may have so little energy loss that for all practical purposes the phase angle may be accepted as 90°. A resistor may be constructed to have a negligible amount of inductance. There is invariably resistance in the wire coil of an inductor and energy loss in the core if it is an iron-core inductor. Hence, its phase angle may deviate considerably from 90°. A circuit containing a nonreactive resistor R in series with a coil having both resistance and reactance is diagrammed in Fig. 5a. The observed voltage VL' across the coil terminals is greater than the reactance drop of the coil VL because of the resistance drop VR, where R is the resistance of the coil wire.
Figure 5: Phase diagram for an ac circuit containing a coil and a resistor in series.
To construct the phase diagram from the observed voltmeter readings, proceed as follows. Using the same scale for VL', VR, and V, draw the horizontal line OA, Fig. 5b, to represent the potential drop VR across the resistor. From the point A describe an arc in the first quadrant of radius VL', the observed potential drop across the coil terminals. From point O describe a second arc of radius V. Construct the lines OB and AB to the point of intersection of the two arcs. Extend the line OA to meet the perpendicular from B. The triangle OBC is the voltage triangle for the circuit. The angle is the lag of current behind VL'. The side BC is the reactance drop of the coil and the side AC is the resistance drop of the coil. When VL, I and f are known, the inductance L of the coil may be computed.
Figure 6: Voltage scale diagram for coil, capacitor, and resistor in series.
When the circuit contains a nonreactive resistor, capacitor and a coil, the phase diagram is constructed as shown in Fig. 6. It is assumed that the capacitor provides a 90° voltage-current phase. The symbols in the phase diagram correspond to the observed voltage readings indicated on the circuit diagram. (An additional voltmeter reading across coil and resistor combined may be used to determine whether or not the assumption of a 90° phase angle for the capacitor is justified.)
Impedance Circuit Apparatus consisting of a Resistor, Inductor, and Capacitor in series; AC ammeter (0.5 amp); AC voltmeter (150/15 volts), high resistance; precision adjustment rheostat.
A switch on the Impedance Circuit Apparatus is arranged to short either the inductor or the capacitor. Thus, a study may be made eliminating either of these elements.
Connect the Ammeter and Rheostat in series with the circuit in the Impedance Circuit Apparatus. Attach to a 120-volt, 60-cycle power supply. (The voltmeter will shunt current across the element to which it is connected. If the resistance of the voltmeter is high this current will be small and a partial compensation is accomplished by adjusting the rheostat for the desired current before each reading of the voltmeter.)
- Close the switch on the Impedance Circuit Apparatus to short out the capacitor. Obtain the voltmeter readings across R, L, and R + L for a current of about 300 ma.
- Close the switch on the Impedance Circuit Apparatus to short out the inductor. Obtain the voltmeter readings across R, C, and R + C for a current of about 300 ma.
- Obtain the voltmeter readings across each element (R, L, and C) and across all elements in series when the current in the circuit is about 300 milliamperes.
GRAPHS, PHASE DIAGRAMS, and CALCULATIONS (Data Analysis):
- Draw the phase diagram using the Data (A). From this diagram determine VL. Compute the inductance value L of the coil.
- Draw the phase diagram using Data (B). From this diagram determine VC . Compute the value of the capacitance C.
- Using Data (C) plot each voltmeter reading against VR as abscissa. What conclusion do you draw from the curves obtained?
- Determine the impedance of the entire circuit. The impedance of the circuit should be equal to the vector sum of the values of the components of the circuit. Check this by appropriate calculations.
Explain to the reader what you are doing. i.e., explain the reasons why the inductance is not pure inductance whereas the capacitor is pure capacitance and the resistor has only resistance. Make sure the reader understands the technique employed.
What can you conclude as a result of doing this experiment? Be frank with the reader. Do not assume she (or he) has had the experience you just had. Tell the why you can say what you can with confidence.
Perhaps you have an idea how we can improve this experiment so that the student can learn easier. This obviously, is optional. Thanks.
- When only the resistor and inductor of the Impedance Circuit Apparatus are used, 90 volts ac is required to provide a current of 0.5 amp. Would the current be the same, greater, or less if 90 volts dc were applied? Explain.
- When 100 volts ac was applied to the three elements (resistor, capacitor and inductor) of the Impedance Circuit Apparatus the current reading was observed as 490 ma. When C was shorted the current rose to 560 ma but when L was shorted it dropped to 290 ma. Show how this is possible using a phase diagram.
- What do the data of problem 2 indicate about the relative value of XL and XC of the Impedance Circuit from which these data were taken?
- A coil having a resistance of 40 ohms and an inductive reactance of 30 ohms is connected to a 120-volt ac source. What is the current and the power in the coil?
- A 60 cycle/sec source is connected first to an inductor and next to a capacitor. In each case the current is 1 amp. What will be the current in the inductor and the current in the capacitor when each is connected to a 600 cycle/sec source of the same voltage?
- A resistor (R = 3 ohms), inductor (XL = 10 ohms) and capacitor (XC = 6 ohms) are connected in series to a 120-volt, ac source. Compute the voltage across the inductor.
- Compute the frequency of a source of voltage V which would give the maximum current when applied to the circuit of the Impedance Apparatus used.
- An inductor takes 600 watts power when connected to a 120-volt, dc source. When it is connected to 120-volts ac, the power dissipated is 120 watts and the observed current is 2 amp. Calculate the actual and the equivalent resistance of the inductor.
- Compute the phase angle for problem 8 and the inductance of the coil.
- Is it possible to have a large current in a circuit connected to a 120-volt, ac line and still dissipate almost no energy in the circuit? Explain.
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