Circuit Analysis in B, Study notes of Basic Electronics

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Basic Electronics Summitted To: Dr. TAQMEEM HUSSAIN Submitted By: MUHAMMAD USMAN (3849) TALHA BASRA (3822) MUHAMMAD HAMZA (3969) GHULAM MOHIUDDIN (3837) MUBASHIR (3859) RUKASH GILL (3838) Semester: 3rd (Morning) Assignment Topic: Circuit Analysis Department: Computer Science Department Of Computer Science Government College University, Faisalabad Table of Contents Circuit Analysis ............................................................................................................................. 1 Basic Elements of Circuit ......................................................................................................... 1 CIRCUIT TERMINOLOGY ....................................................................................................... 2 Ohm’s Law .................................................................................................................................... 2 Ways to Connect Circuit Components ........................................................................................ 3 Series Circuit ............................................................................................................................. 4 Parallel Circuit .......................................................................................................................... 5 Voltage Divider.............................................................................................................................. 7 Current Divider ............................................................................................................................. 8 Kirchhoff’s Law .......................................................................................................................... 10 Kirchhoff’s Current Law........................................................................................................ 10 Kirchhoff’s Voltage Law ........................................................................................................ 12 Network Theorems...................................................................................................................... 13 Superposition Theorem........................................................................................................... 14 Thevenin’s Theorem ............................................................................................................... 14 Norton’s Theorem ................................................................................................................... 14 Thevenin’s And Norton’s Theorem Equivalence ..................................................................... 14 3 Formula: 𝐼 ∝ 𝑉 𝐼 = ( 1 𝑅 ) ∗ 𝑉 Purpose of Ohm’s law in Circuit Analysis: Ohm's Law is a formula used to calculate the relationship between voltage, current, and resistance in an electrical circuit. Numerical No. 1: Find the current across the given circuit? Solution: 𝐼 = ( 1 𝑅 ) ∗ 𝑉 𝑰 = ( 𝟏 𝟏𝟎𝟎 ) ∗ 𝟗 𝑰 = 𝟎. 𝟎𝟗𝑨 The current across the circuit is 0.09A. Ways to Connect Circuit Components: There are two basic ways in which to connect more than two circuit components: 1. Series Circuit 2. Parallel Circuit Here, 1/R is a constant of proportionality known as conductance. 4 Series Circuit: When components in a circuit are connected end-to-end so that all the circuit current passes through each component, they form a series circuit. In the above circuit, there are three resistors of resistance R1, R2, and R3 connected in series with each other and the battery. This results in only one path for the current flow. Hence, current I am the same in all three resistors. Due to this current flow, voltages drop V1, V2 and V3 occur across R1, R2, and R3 respectively. Therefore, V1=IR1 V2=IR2 V3=IR3 The sum of the voltages will be: Veq= V1+ V2+ V3 IReq= IR1+ IR2+ IR3 Req= R1+R2+R3 Equivalent resistance for n number of resistors in a circuit: Req= R1+R2+R3+…. +Rn Numerical No. 2: A series circuit has resistance R1, R2, and R3 of 17Ω,12 Ω and 11 Ω and total voltages across the circuit are 60V. Find the current flowing through the circuit? 5 Solution: Req= R1+R2+R3 Req=17+12+11 Req=40Ω Veq=IReq 60=I*40 I=1.5A So, the current in the circuit will be 1.5A. Parallel Circuit: A parallel is a branched arrangement in which two or more resistors are connected side by side across a single voltage source. In the above circuit, there are two resistors of resistance R1, and R2 connected in a parallel way with each other and the battery. This results in only one path for the current flow. Hence, voltage V is the same in all three resistors. Due to this current flow, the current flowing through the resistances will vary such as I1, and I2 occurring across R1, and R2, respectively. Therefore, I1 = V/R1 I2 = V/R2 The sum of the voltages will be: Ieq= I1+ I2 V/Req= V/R1+ V/R2 8 Solution: Req = R₁ + R₂ = 40 + 60 = 100Ω I = V/Req= 12*100 = 0.12A or 120mA V1 = (V/(R1+R2)) * R1= (12/100) *40=4.8V V2 = (V/(R1+R2)) * R2= (12/100) *60=7.2V Current Divider: Current Divider circuits have two or more parallel branches for currents to flow through, but the voltage is the same for all components in the parallel circuit. • It is a parallel circuit where the output current is observed in one of the branches. Suppose we must find the current flowing through the resistance R1, the current will be IR1 and will be: IR1=Vs/R1 ------(1) Finding the Vs, we use equivalent resistance formula for the parallel circuit: 1/Req= 1/R1+1/R2 Req=[1/R1+1/R2] -1 -----(2) As we know Req=Vs/IT, here IT is the total current passing through a circuit. By putting value of Req in equation (2): Vs/IT=[1/R1+1/R2] -1 Vs=IT [ (R1+R2)/R1R2] -1 Vs=IT [ R1R2/(R1+R2)] ------(3) 9 Now, putting equation (3) in equation (1): IR1=IT [ R1R2/(R1+R2)] *(1/R1) IR1=IT[R2/(R1+R2)] Similarly, IR2=IT[R1/(R1+R2)] The above equations for each branch current have the opposite resistor in its numerator. That is to solve for I1 we use R2, and to solve for I2 we use R1. This is because each branch current is inversely proportional to its resistance resulting in the smaller resistance having the larger current. Numerical No. 5 Two parallel resistors having their values 50 ohms and 100 ohms are connected in parallel. Find the current flowing through each when the connected source is 20 A. Solution: IR1=IT[R2/(R1+R2)] IR1=20[100/ (50+100)] IR1=13.33A IR2=IT[R1/(R1+R2)] IR2=20[50/ (50+100)] 10 IR2=20[0.33] IR2=6.66A Kirchhoff’s Law: Circuit with two or more batteries connected in its different branches are not explained by parallel or series circuit means rules of parallel and series circuits are incapable for these circuits. These circuits can be easily solved with the help of Kirchhoff’s law which is given below: Kirchhoff’s Current Law: It states that in any network of conductors, the algebraic sum of currents meeting at a point (or junction) is zero. OR The sum of all the current in a circuit is equal to zero. Mathematical representation: ∑𝑰 = 𝟎 Applying KCL at node given in figure: Sum of entering current = Sum of exiting current I1+I2+=I3+I4 13 Numerical No. 7 In the circuit below e1=20, VR1=5Volts and e2=10Volts.Find the voltages VR2 and VR3. Solution: Apply Kirchhoff's law of voltage to loop L1 and the equation will be: e1−VR1−VR2=0 Putting the values of e1 and VR2 we have, 20−5−VR2=0 VR2= 15Volts Apply Kirchhoff's law of voltage to loop L2and the will equation be: VR2+e2−VR3=0 Putting the vales of VR2 and e2 we have, 15+10−VR3=0 VR3= 25Volts Network Theorems: As we already discussed about most of networks cannot be solved by applying law of parallel and series circuits. Therefore, we use Kirchhoff laws, but it often makes the solution long and laborious. Hence, various networks theorems have developed which provide very short and time-saving methods to solve these complicated circuits. Here, we discuss three networks’ theorems are given below: 14 1. Superposition Theorem 2. Thevenin’s Theorem 3. Norton’s Theorem Superposition Theorem: The superposition theorem states that a circuit with multiple voltage and current sources is equal to the sum of simplified circuits using just one of the sources. Superposition says that a circuit with multiple sources can be solved by this process: 1. Set all sources = 0, except one. 2. Solve necessary currents and voltages, using only that source. 3. Repeat Step 2 for each source. 4. Finally, algebraic sum of currents and voltage drops over a resistor due to different sources is taken. It gives the actual value of current, and voltage drop in that resistor or component. Thevenin’s Theorem: Thevenin's theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. The basic procedure for solving a circuit using Thevenin’s Theorem is as follows: 1. Remove the load resistor RL or component concerned. 2. Find RS by shorting all voltage sources or by open circuiting all the current sources. 3. Find VS by the usual circuit analysis methods. 4. Find the current flowing through the load resistor RL. Norton’s Theorem: Norton's theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. The basic procedure for solving a circuit using Norton’s Theorem is as follows: 1. Remove the load resistor RL or component concerned. 2. Find RS by shorting all voltage sources or by open circuiting all the current sources. 3. Find IS by placing a shorting link on the output terminals A and B. 4. Find the current flowing through the load resistor RL. Thevenin’s And Norton’s Theorem Equivalence: Since Thevenin’s and Norton’s Theorems are two equally valid methods of reducing a complex network down to something simpler to analyze, there must be some way to convert a Thevenin equivalent circuit to a Norton equivalent circuit, and vice versa.