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This document delves into the fundamentals of electrical engineering, specifically focusing on ac and dc circuit analysis. it covers key concepts such as resistance, impedance, and network theorems like kirchhoff's laws, superposition, thevenin's theorem, and norton's theorem. the document uses numerous examples to illustrate calculations and problem-solving techniques related to various circuit configurations, including series, parallel, and series-parallel arrangements. it also explores the application of delta-star and star-delta transformations for circuit simplification. The educational value lies in its practical approach, providing step-by-step solutions and detailed explanations of complex electrical concepts.
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An ammeter is an instrument used to measure current and must be connected in series with the circuit. Since all the current in the circuit passes through the ammeter it must have a very low resistance. A voltmeter is an instrument used to measure p.d. and must be connected in parallel with the part of the circuit whose p.d. is required. To avoid a significant current flowing through it a voltmeter must have a very high resistance. An ohmmeter is an instrument for measuring resistance. A multimeter , or universal instrument, may be used to measure voltage, current and resistance. Multiples and sub-multiples Prefix Name Meaning G Giga Multiply by 1,000,000,000 (i.e. X 109 ) M Mega Multiply by 1,000,000(i.e. X 10^6 ) K Kilo Multiply by 1,000 (i.e. X 10^3 ) m mili Divide by 1,000 (i.e. X 10-^3 ) μ micro Divide by 1,000,000 (i.e. X 10-^6 )
RESISTANCE It may be defined as the property of a substance due to which it opposes (or restricts) the flow of electricity (current) through it. The resistance of an electrical conductor depends on the following factors; (a) Varies directly as its length l (if the length of a piece of wire is doubled, then the resistance is doubled).
(a) Resistance, R , is directly proportional to length, l , i.e. 𝑅 ∝ 𝑙. Hence, 600 Ω ∝ 5m or 600=( k )(5), where k is the coefficient of proportionality. Hence, 𝑘 = 600/5 = 120 When the length l is 8 m, then resistance 𝑹 = 𝑘𝑙 = (120)(8) = 𝟗𝟔𝟎 Ω (b) When the resistance is 420 Ω , 420= kl , from which, length 𝒍 = 420/𝑘 = 420/120 = 𝟑. 𝟓𝒎
Example 2: A piece of wire of cross-sectional area 2mm2 has a resistance of 300 Ω. Find; (a) The resistance of a wire of the same length and material if the cross-sectional area is 5mm^2 , (b) The cross-sectional area of a wire of the same length and material of resistance 750 Ω. Solution Resistance R is inversely proportional to cross-sectional area, A , i.e. 𝑅 ∝ 𝑙/𝐴
Hence, 300 Ω ∝ 12 𝑚𝑚^2 , or 300 = (𝑘)(^12 ) from which, the coefficient of
proportionality, 𝑘 = 300 × 2 = 600 (a) When the cross-sectional area a =5mm^2 then
𝑅 = (𝑘)(^15 ) = (600)(^15 ) = 𝟏𝟐𝟎 Ω (Note that resistance has decreased as the cross-sectional is increased.)
(b) When the resistance is 750 Ω then 750 Ω = (𝑘) (^) 𝐴^1
from which cross-sectional area,
𝐴 = 750 𝑘 =^600750 = 𝟎. 𝟖𝒎𝒎𝟐
Temperature coefficient of resistance In general, as the temperature of a material increases, most conductors increase in resistance, insulators decrease in resistance, while the resistance of some special alloys remains almost constant. The temperature coefficient of resistance of a material is the increase in the resistance of a 1 Ω resistor of that material when it is subjected to a rise of temperature of 1⁰C. The symbol used for the temperature coefficient of resistance is α (Greek alpha). If the resistance of a material at 0⁰C is known the resistance at any other temperature can be determined from: 𝑹𝜽 = 𝑹𝟎(𝟏 + 𝜶𝟎𝜽) Where; 𝑅 0 = resistance at 0⁰C 𝑅𝜃 = resistance at temperature 𝜃°𝐶 𝛼 0 = temperature coefficient of resistance at 0⁰C Example 3. A coil of copper wire has a resistance of 100 Ω when its temperature is 0⁰C. Determine its resistance at 70⁰C if the temperature coefficient of resistance of copper at 0⁰C is 0.0043 / ⁰C. Resistance 𝑅𝜃 = 𝑅 0 (1 + 𝛼 0 𝜃) Hence resistance at 100⁰C, 𝑅70 = 100[1 + (0.0043)(70)] = 100[1 + 0.301] = 100(1.301) = 100(1.301) = 𝟏𝟑𝟎. 𝟏 Ω Example 4. A carbon resistor has a resistance of 1 K Ω at 0⁰C. Determine its resistance at 80⁰C. Assume that the temperature coefficient of resistance for carbon at 0◦C is − 0_._ 0005 / ⁰C. Resistance at temperature 𝜃 ⁰C, 𝑅𝜃 = 𝑅 0 (1 + 𝛼 0 𝜃) 𝑅𝜃 = 1000[1 + (−0.0005)(80)]
i.e. V = V 1 + V 2 + V 3 From Ohm’s law: V 1 = IR 1 , V 2 = IR 2 , V 3 = IR 3 and V = IR where R is the total circuit resistance. Since V = V 1 + V 2 + V 3 then IR = IR 1 + IR 2 + IR 3. Dividing throughout by I gives R = R 1 + R 2 + R 3 Thus for a series circuit, the total resistance is obtained by adding together the values of the separate resistance’s. The main characteristics of series circuits Same currents flows through all parts of the circuit Different resistors have different voltage drops Voltage drops are additive Applied voltage equall the sum of different voltage drops Resistances are additive Powers are additive
Example 1 For the circuit below determine; (a) The battery voltage V , (b) The total resistance of the circuit, (c) The values of resistors R 1 , R 2 and R 3 , given that the p.d.’s across R 1, R 2 and R 3 are 5V, 2V and 6V respectively.
(a) Battery voltage 𝑉 = 𝑉 1 + 𝑉 2 + 𝑉 3 , = 5 + 2 + 6 = 𝟏𝟑𝑽
(b) Total circuit resistance 𝑅 = 𝑉𝐼 = 134 = 𝟑. 𝟐𝟓𝛀 (c) Resistance 𝑅 = 𝑉𝐼 = 54 = 𝟏. 𝟐𝟓𝛀 Resistance 𝑅 = 𝑉𝐼 = 24 = 𝟎. 𝟓𝛀 Resistance 𝑅 = 𝑉𝐼 = 64 = 𝟏. 𝟓𝛀 (Check: R1 +R2 +R3 =1.25+0.5+1.5 = 3.25Ω = R)
Potential Divider The voltage distribution for the circuit shown in Fig. (a) is given by:
𝑉 1 = (^) 𝑅 𝑅^1 1 +^ 𝑅 2
The circuit shown in Fig. 5.5(b) is often referred to as a potential divider circuit. Such a circuit can consist of a number of similar elements in series connected across a voltage source, voltages being taken from connections between the elements. Frequently the divider consists of two resistors as shown in Fig. (b), where;
𝑉𝑂𝑈𝑇 = (^) 𝑅 1 𝑅+^2 𝑅 2 𝑉𝐼𝑁
Value of unknown resistance; 𝑅𝑥 = 8 − 2 = 𝟔 Ω (b) P.d. across 2 Ω resistor; 𝑉1 = 𝐼 𝑅 1 = 3 × 2 = 𝟔𝑽 Alternatively, from above;
𝑉 1 = (^) 𝑅 1 𝑅+^1 𝑅𝑥 𝑉, = (^) 2 + 2^2 24 = 𝟔𝑽
Energy used = power × time = 𝑉 × 𝐼 × 𝑡 = ( 24 × 3 𝑊)( 50 ) = 3600 𝑊 = 𝟑. 𝟔𝑲𝑾𝒉
DC Parallel Circuits Figure below shows three resistors, R 1, R 2 and R 3 connected across each other, i.e. in parallel, across a battery source of V volts.
In a parallel circuit: (a) the sum of the currents I 1 , I 2 and I 3 is equal to the total circuit current, I , i.e. 𝑰 = 𝑰𝟏 + 𝑰 (^) 𝟐 + 𝑰𝟑
(b) the source p.d., V volts, is the same across each of the resistors. From Ohm’s law:
𝐼 1 = (^) 𝑅𝑉 1 , 𝐼 2 = (^) 𝑅𝑉 2 , 𝐼 3 = (^) 𝑅𝑉 3 𝑎𝑛𝑑 𝐼 = 𝑉𝑅
Where R is the total circuit resistance. Since
𝑰 = 𝑰𝟏 + 𝑰 (^) 𝟐 + 𝑰𝟑 𝑡𝑒𝑛 𝑉𝑅 = (^) 𝑅𝑉 1
2
3 Dividing through by V gives 1 𝑅 =
This equation must be used when finding the total resistance R of a parallel circuit. Considering two resistors in parallel 1 𝑅 =
Hence
𝑅 = (^) 𝑅𝑅 11 +𝑅𝑅^22 𝑖. 𝑒. 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠𝑢𝑚
The main characteristics of parallel circuits Same voltage acts at all points of the circuit Different resistors have different currents flowing through them Branch currents are additive Powers are additive Example 1****. Two resistors, of resistance 3 Ω and6 Ω, are connected in parallel across a battery having a voltage of 12V. Determine; (a) The total circuit resistance (b) The current flowing in the 3 Ω resistor. The circuit diagram is shown;
Or
𝑅 = 6Ω, 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 = 𝑉𝑅 =^606 = 𝟏𝟎𝑨
Current Division For the circuit shown, the total circuit resistance, R T is given by;
𝑅𝑇 = (^) 𝑅𝑅 11 +𝑅𝑅^22
and 𝑉 = 𝐼𝑅𝑇 = 𝐼 (^) 𝑅𝑅 11 +𝑅𝑅^22
i.e.
𝑰𝟏 = (^) 𝑹𝟏𝑹+𝟐𝑹𝟐 (𝑰) 𝒂𝒏𝒅 𝑰𝟐 = (^) 𝑹𝟏𝑹+𝟏𝑹𝟐 (𝑰)
Example 1: For the series-parallel arrangement shown, find (a) The supply current, (b) The current flowing through each resistor and (c) The p.d. across each resistor.
Solution (a) The equivalent resistance R x of R 2 and R 3 in parallel is: 𝑅 2 𝑅 3 𝑅 2 +𝑅 3 =
The equivalent resistance 𝑅𝑇 𝑜𝑓 𝑅 1 , 𝑅𝑥 𝑎𝑛𝑑 𝑅 4 in series is; 𝑅𝑇 = 2.5 + 1.5 + 4 = 8Ω Supply current; 𝑉 𝑅𝑇^ =
(b) The current flowing through 𝑅 1 𝑎𝑛𝑑 𝑅 4 is 25A Current flowing through 𝑅 2 = (^) 𝑅 2 𝑅+^3 𝑅 3 𝐼 , = (^) 6 + 2^2 25 = 𝟔. 𝟐𝟓𝑨
𝑅 3 = (^) 𝑅 2 𝑅+^2 𝑅 3 𝐼 , = (^) 6 + 2^6 25 = 𝟏𝟖. 𝟕𝟓𝑨 NB: Currents flowing through 𝑅 2 𝑎𝑛𝑑 𝑅 3 must add up to the total current flowing into the parallel arrangement i.e. 25A. (c) The equivalent circuit of the circuit above is as shown;
Voltage across 𝑅 1 , 𝑖. 𝑒. , 𝑉 1 = 𝐼𝑅 1 = 25 2.5 = 𝟔𝟐. 𝟓𝑽
i. Each lamp has 240V across it and thus each will glow brilliantly at their rated voltage. ii. If any lamp is removed from the circuit or develops a fault (open circuit) or a switch is opened, the remaining lamps are unaffected. iii. The addition of further similar lamps in parallel does not affect the brightness of the other lamps. iv. More cable is required for parallel connection than for a series one. The parallel connection of lamps is the most widely used in electrical installations. Short and Open Circuits When two points are connected together by a thick metallic wire (Fig. a), they are said to be short-circuited. Since ‘short’ has practically zero resistance, it gives rise to two important facts; i. No voltage can exist across it because V=IR=1 x 0= ii. Current through it (called short-circuit current) is very large (theoretically, infinity)
Two points are said to be open-circuited when there is no direct connection between them (Fig. b). An ‘open’ represents a break in the continuity of the circuit. Due to this break; i. Resistance between the two points is infinite ii. There is no flow of current between the two points
There are Two General Approaches to Network Analysis i. Direct method: In this method, the network is left in its original form while determining its different voltages and currents. Such methods are restricted to fairly simple circuits and include Kirchhoff’s and superposition. ii. Network reduction method: In this method the original network is converted into a much smaller equivalent circuit for rapid calculation of different quantities. This method can be applied to simple as well as complicated networks. They include Delta/star and Star/Delta conversions, Thevenin’s theorem and Norton’s theorem
Kirchhoff’s laws Kirchhoff’s laws state Current Law. At any junction in an electric circuit the total current flowing towards that junction is equal to the total current flowing away from the junction, i.e. Σ𝐼 = 0_._ Thus, referring to the figure below
I 1 + I 2 = I 3 + I 4 + I 5 or I 1 + I 2 − I 3 − I 4 − I 5 = Voltage Law: In any closed loop in a network, the algebraic sum of the voltage drops (i.e. products of current and resistance) taken around the loop is equal to the resultant e.m.f. acting in that loop or The algebraic sum of the products of currents and resistance in each of the conductors in any closed path in a network plus the algebraic sum of the e.m.fs in that path is zero. Thus, referring to the figure below;
(Note that if current flows away from the positive terminal of a source, that source is considered by convention to be positive. Thus moving anticlockwise around the loop of figure above, E 1 is positive and E 2 is negative).
Example 1: (a) Find the unknown currents marked in Fig. (a). (a) Determine the value of e.m.f. E in Fig. (b).
Solution (a) Applying Kirchhoff’s current law: For junction B: 50=20+ I 1. Hence I 1 =30A For junction C: 20+1 5 = I 2 .Hence I 2 =35A For junction D: I 1 = I 3 +120 i.e. 30= I 3 +120. Hence I 3 =−90A (i.e. in the opposite direction to that shown in Fig. (a)) For junction E: I 4 + I 3 =1 5 i.e. I 4 =1 5 −(−90). Hence I 4 =105A For junction F: 120= I 5 +40. Hence I 5 =80A (b) Applying Kirchhoff’s voltage law and moving clockwise around the loop of Fig. (b) starting at point A: 3 + 6 + E − 4 = ( I )(2) + ( I )(2_._ 5) + ( I )(1_._ 5) + ( I )(1) = I (2 + 2_._ 5 + 1_._ 5 + 1)
