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Electrical Calculations
Learning Outcome
When you complete this learning material, you will be able to:
Define terms and perform simple calculations involving DC and AC power circuits.
Learning Objectives
You will specifically be able to complete the following tasks:
- Use Ohm’s Law and Kirchhoff’s Laws to calculate current, resistance or voltage drop in series or parallel multi-resistor circuits.
- Calculate unknown resistances using a Wheatstone Bridge circuit.
- Explain and perform calculations involving electrical power, work and energy.
- Calculate the frequency, period and phase angle for an ac sine wave.
- Define terms and calculate the peak-to-peak, root mean square, and maximum values for ac voltage and current.
- Given required parameters, calculate the inductive reactance, capacitive reactance, total reactance, and impedance for an ac circuit, plus circuit frequency and current flow.
- Calculate real power, imaginary power and power factor for an ac circuit.
- Given the load, voltage and power factor of a 3-phase generator, calculate the kVA and kW ratings of the generator.
Objective One
When you complete this objective you will be able to…
Use Ohm’s Law and Kirchhoff’s Laws to calculate current, resistance or voltage drop in series or parallel multi-resistor circuits.
Learning Material
UNITS USED FOR CALCULATIONS
The potential difference (p.d.) is the difference in electric charge between two points. P.d. is measured in volts. A device that has the ability to maintain a potential difference in charge between two points is said to develop an electromotive force (emf). A potential difference causes a current to flow and an emf maintains a potential difference. Both are measured in volts. As both are measured in volts, a common term, voltage, is used to indicate a measure of either. Although potential difference, emf, and Voltage do not mean exactly the same thing, they are often used interchangeably. In calculations, E or V are used for voltage, emf, and potential difference.
Simple Direct Current Circuits At its simplest, an electric circuit has a source of electromotive force, a wire or conductor connecting the source to a load or resistance, and a second wire connecting the load to the source again. (See Fig. 1). Fig. 1 shows an electric circuit where I is the current in amperes (A), E is the electromotive force in volts (V), and R is the resistance in ohms (Ω). Figure 1 Simple Direct Current Circuit OHM’S LAW Ohm’s Law states the relationship between current, potential difference (change in voltage), and resistance as found by experiment. It states that current is directly proportional to electromotive force and inversely proportional to resistance. It can be written as: KIRCHHOFF’S LAWS More complicated electric circuits may be solved with the aid of two simple rules, Kirchhoff’s laws. First Law – Kirchhoff’s Current Law
Figure 3 Change of Voltages Figure 4 Current Flow Example 1: Write Kirchhoff’s laws for all the current junctions and closed paths in Fig. 5. Figure 5 Solution: Show the possible current paths and indicate the direction of all changes in potential. Current Law for junction A: I 1 = I 2 + I 3 Current Law for junction B: I 1 = I 2 + I 3
Figure 6 Path 1: V 1 + V 2 + (-E 1 ) = 0 I 1 R 1 + I 2 R 2 + (- E 1 ) = 0 (Ans.) Path 2: -E 2 + V 3 + (-V 2 ) = 0 (-E 2 ) + I 3 R 3 + (-I 2 R 2 ) = 0 (Ans.) Path 3: V 1 + (-E 2 ) + V 3 + (-E 1 ) = 0 I 1 R 1 + (-E 2 ) + I 3 R 3 + (-E 1 ) = 0(Ans.) Series Circuits With resistors connected in series, the same current flows through all the resistors. The total resistance is equal to the sum of all the resistances in the series. The sum of the voltage drops across the resistors in the series is equal to the total potential (voltage) drop in the circuit. The supplied emf in volts is also equal to the total potential drop in the circuit.
The voltage drop across R1 equals V1 and: V 1 = IR 1 The voltage drop across R2 equals V2 and: V 2 = IR 2 The voltage drop across R3 equals V3 and: V 3 = IR 3 Finally the voltage drop, Vt, across all the resistors equals: Vt = V 1 + V 2 + V 3 Substituting IR 1 for V 1 , IR 2 for V 2 and IR 3 for V 3 , the equation is: Vt = IR 1 + IR 2 + IR 3 or Vt = I(R 1 + R 2 + R 3 ) Example 2: Fig. 8 shows a series circuit with three resistors. Using Ohm’s Law calculate: (a) The total resistance in the circuit. (b) The voltage drop across each resistor. (c) The total voltage drop.
Figure 8 Series Circuit Solution: (a) The total resistance in the circuit equals Rt. Rt = R 1 + R 2 + R 3 = 4 Ω + 6 Ω + 8 Ω = 18 Ω (Ans.) (b) The voltage drop across R 1 equals V 1. V 1 = IR 1 = 2 A x 4 Ω = 8 V (Ans.) The voltage drop across R 2 equals V 2. V 2 = IR 2 = 2 A x 6 Ω = 12 V (Ans.)
Figure 9 Series Circuit Solution : (a) The total resistance equals Rt. Rt = R 1 + R 2 + R 3 + R 4 = 5 Ω + 4 Ω + 10 Ω + 6 Ω = 25 Ω (Ans.) (b) The current in the circuit equals I. I = E/Rt = 50 V/25 Ω = 2 A (Ans.) (c) The voltage drop across R 1 equals V 1. V 1 = IR 1 = 2 A x 5 Ω = 10 V (Ans.) The voltage drop across R 2 equals V 2.
V 2 = IR 2
= 2 A x 4 Ω = 8 V (Ans.) The voltage drop across R 3 equals V 3. V 3 = IR 3 = 2 A x 10 Ω = 20 V (Ans.) The voltage drop across R 4 equals V 4. V 4 = IR 4 = 2 A x 6 Ω = 12 V (Ans.) Checking the total of all voltage drops, or the applied emf (E), will provide a quick check of your answer: E = V 1 + V 2 + V 3 + V 4 = 10 V + 8 V + 20 V + 12 V = 50 V Example 4 : The series circuit shown in Fig.10 consists of four known and one unknown resistance. The resistors are supplied with a current of 5 A from a generator producing an applied emf of 110 V. Calculate: (a) The total resistance in the circuit. (b) The resistance of R5. (c) The voltage drop across each resistor.
and: V 2 = IR 2 = 5 A x 5 Ω = 25 V (Ans.) and: V 3 = IR 3 = 5 A x 2 Ω = 10 V (Ans.) and: V 4 = IR 4 = 5 A x 8 Ω = 40 V (Ans.) and: V 5 = IR 5 = 5 A x 4 Ω = 20 V (Ans.) Checking for Vt which equals the applied emf (E): E = V 1 + V 2 + V 3 + V 4 +V 5 = 15 V + 25 V + 10 V + 40 V + 20 V = 110 V Parallel Circuits Fig. 11(a) shows a circuit with resistors in parallel. In a parallel circuit, the sum of the individual currents through each loop or path is equal to the total current in the circuit. This is different from a series circuit where the same current flows through all the resistors. In a parallel circuit the same voltage is applied to all resistors. Fig. 11(b) shows another way to represent a parallel circuit that may make the circuit easier to
visualize. Figure 11 Parallel Circuits The total resistance of the circuit in Fig. 11 is: Rt = E/It The total current is: It = I 1 + I 2 + I 3 and the voltage is: E = I 1 R 1 or E = I 2 R 2 or E = I 3 R 3 When Ohm’s Law is applied to the individual resistors, the individual currents can be expressed as: I 1 = E/R 1 I 2 = E/R 2
Figure 12 Parallel Circuit Solution: (a) The equivalent resistance equals Rt. 1/Rt = (1/R 1 ) + (1/R 2 ) + (1/R 3 ) + (1/R 4 ) 1/Rt = (1/5 Ω) + (1/8 Ω) + (1/6 Ω) + (1/16 Ω) 1/Rt = (0.200 Ω) + (0.125 Ω) + (0.167 Ω) + (0.062 Ω) 1/Rt = 0.554 Ω Rt = 1 Ω/0. = 1.805 Ω (Ans.) (b) The total current equals It. It = E/Rt = 100 V/1.805 Ω
= 55.40 A (Ans.) (c) The current through R 1 equals I 1. I 1 = E/R 1 = 100 V/5 Ω = 20 A (Ans.) The current through R 2 equals I 2. I 2 = E/R 2 = 100 V/8 Ω = 12.5 A (Ans.) The current through R 3 equals I 3. I 3 = E/R 3 = 100 V/6 Ω = 16.67 A (Ans.) The current through R 4 equals I 4. I 4 = E/R 4 = 100 V/16 Ω = 6.25 A (Ans.) Check with: It = I 1 + I 2 + I 3 + I 4 It = 20 A + 12.5 A + 16.67 A + 6.25 A = 55.42 A (The slight difference between the calculated values is due to using fractions in one method and decimals in the other method.)
(b) The current through R 1 is I 1 = E/R 1 = 120 V/10 Ω = 12 A (Ans.) The current through R 2 is I 2 = E/R 2 = 120 V/12 Ω = 10 A (Ans.) The current through R 3 is I 3 = E/R 3 = 120 V/15 Ω = 8 A (Ans.) Check with: It = I 1 + I 2 + I 2 = 12 A + 10 A + 8 A = 30 A Example 7: Calculate the resistance, R 4 , for the circuit shown in Fig. 14.
Figure 14 Parallel Circuit Solution: This and many problems can often be solved by different methods. Two different approaches are used and shown in this example. Method 1: From the information we can calculate the total resistance, Rt. Rt = E/It = 120 V/45 A = 2.67 Ω and: 1/Rt = (1/R 1 ) + (1/R 2 ) + (1/R 3 ) + (1/R 4 ) 1/2.67 Ω = (1/15 Ω) + (1/8 Ω) + (1/10 Ω) + (1/R 4 ) 0.375/Ω = 0.067/Ω + 0.125/Ω + 0.100/Ω + 1/R 4
