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This Physics project aims to understand the scalar nature of current, charge, EMF, and P.D., and how to apply Kirchoff's laws to complex electrical networks. Learn about sign conventions and how to calculate current values at different branches.
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WELCOME TO PHYSICS PROJECT
OBJECTIVES To understand that current, charge, emf and P.D are all scalars and can be added algebraically. To apply Kirchoff’s laws to complex electrical network to find the values of current at different branches. To draw directions of currents and to add currents and voltage drops using sign conventions.
PREVIOUS KNOWLEDGE 1.Ohm’s law : V =IR , at constant temperature. 2.Current in series: the same amount of current will flow across the resistors in series. 3.Voltage in parallel: the same voltage is been maintained across the resistances in parallel.
EMF is the potential difference between any two points in a circuit when no current is drawn from the circuit. • Potential difference is the voltage across any two points in a circuit when a current is drawn from the circuit.
6.Both emf and p.d are measured in Volts. 7. For a unit charge, Both emf and p.d are synonymous to energy .
Use of Kirchoff’s laws • Ohm’s law can be applied to simple circuits only .For complex electrical network, where current gets branched , only Kirchoff’s laws are effective.
Some important terms • Circuit: It is a network consisting of electrical elements & devices through which current is passed
Open circuit : It is an electrical circuit which has a discontinuity and hence current does not pass through it.
Closed circuit : It is an electrical circuit which is continuous and closed so that a current passes through it.
Junction : It is a point on an electrical circuit where currents from different paths meet.
KIRCHOFF’SFIRST LAW • This law is also called Kirchoff’s current law or Junction rule
SIGN CONVENTIONS • All currents flowing towards the the junction are taken positive. • All currents flowing away from the junction are taken negative.
STATEMENT The algebraic sum of all currents meeting at a point or a junction in an electrical network is zero.
Applying sign convention to the sample diagram we geti1-i2 –i3=0
Physical meaning of KCL • It is the consequence of law of conservation of current.
Hydraulic analogy :when there is no source or sink inside pipes the amount of water entering into the pipes is equal to the amount of water leaving the pipes .
KIRCHOFF’S SECOND LAW • This law is also called Kirchoff’s Voltage law or Loop rule. • This law is a consequence of the law of conservation of energy.
Sign conventions As one traverse a closed loop. 1.Voltage drop flowing in the direction of decreasing potentials are taken as - ve & those in the direction of increasing potentials are taken + ve. continued
2.EMF flowing in the direction of decreasing potentials are taken as - ve & those in the direction of increasing potentials are taken + ve.
STEPS TO FRAME EQUATIONS • STEP 1. Choose a particular closed path . Start travelling through the selected path cyclically either in a clockwise or anti clockwise direction.
STEP 2 Apply KVL for the loop and record the voltage drops and the emf in the ckt as per sign conventions.
STEP 3 The number of equations to be formed must be equal to the number of unknown currents. The equations so obtained will be simultaneous equations
STEP 4 solve the simultaneous equations and the values of unknown currents are to be found.
Applying the KVL for the sample diagram , Consider the closed loop ABEFA -I1R1-( I1+I2)R3-E1 = 0 -------- (1) Consider the closed loop BCDEB I2R2 +E2 +(I1+I2)R3 = 0-----(2)
Solving equations (1) & (2), the values of I1 and I2 are calculated.
EXAMPLE • Find the voltage drop across 5 ohm resistor for the following network.
Given R1= 1Ω , R2 = 2 Ω, R3= 5 Ω E1 = 2v E2 = 5v consider the closed loop ABEFA, Applying KVL, we get -I1+-5(I1+I2) – 2 = 0 -6 I1-5 I2 =2 ---------------- (1)
Applying KVL for BCDEB, We get 2 I2 +5 + 5 ( I1 +I2 ) = 0 5 I1 + 7 I2 = -5 ---------------------- (2) Solving equations (1) & (2) ,we get I1 = 11/17 A I2 = -20/17 A
Current through 5 ohm resistor is I = I1 + I2 I = 11/17 -20/17 I = -9/17 A So p.d across 5 ohm resistor is V = I x R V = -9/17 x 5= 45/17 Volts. ( Neglecting the negative sign)
TEST YOUR KNOWLEDGE • Kirchoff’s first law is a consequence of the law of conservation of a) current b) voltage c) energy
Kirchoff’s voltage law is also called a) Loop rule b) Networking c) Junction rule
WEBSITES FOR REFERENCE • www.physicsclassroom.com • www.sasked.gov.sk.ca • www.bookpool.com • www.online.physics.uiuc.edu