BEEE DC 1 PDF

Circuit Elements Electric Charge and Current  The most basic quantity in an electric circuit is the electric charge.  We all experience the effect of electric charge when we try to remove our wool sweater and have it stick to our body or walk across a carpet and receive a shock. Charge is an...

Circuit Elements Electric Charge and Current  The most basic quantity in an electric circuit is the electric charge.  We all experience the effect of electric charge when we try to remove our wool sweater and have it stick to our body or walk across a carpet and receive a shock. Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C). 1 1 Circuit Elements What is an electric circuit?  It is a closed path for transmitting an electric current through the medium of electrical and magnetic fields.  The flow of electrons across the loop constitutes the electric current.  Electrons enter the circuit through the ‘Source’ which can be a battery or a generator. 3 2 Circuit Elements  Passive Element: The element which receives energy (or absorbs energy) and then either converts it into heat (R) or stored it in an electric (C) or magnetic (L ) field is called passive element. 6 3 4 5 6 Circuit Elements: Resistor Resistance (R)  Resistance is a “property” of material which opposes the flow of electricity through it when potential is applied. 13 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25  From ohm’s law  V1 = 2 x i V2 = 3 x i _____________(1)  Applying KVL around the loop gives  20 - V1 - V2 = 0 _________________(2)  Substituting Eq.(1) into Eq.(2)  20 - 2 i– 3 i = 0 or 5 i = 20 therefore, i = 4 A.  Substituting I in Eq.(1) finally gives  V1 = 2 x i = 8 V and V2 = 3 x i = 12 V. 26 Kirchhoff’s Laws Observations from the Circuit Junction: A & D Loop: ABDA & ACDBA Current follows the Ohm’s law 53 27  We will apply Ohm’s Law and Kirchhoff’s Laws. By Ohm’s Law,  V1 = 8 x I1 V2 = 3 x I2 V3 = 6 x I3  Applying KCL at node a,  i1 – i2 – i3 =0 _______________(1)  Applying KVL to Loop 1,  30 – V1 – V2 = 0 _______________(2)  We express Eq.(2) in terms of i1 and i2  30 – ( 8 x I1) – (3 x I2 ) = 0________(3)  Therefore, i1 = [30 – (3 x I2 )] / 8 28  30 – 3 I2 – 8 i2 – 4 i2 = 0  30 – 15 I2 = 0  Which gives the value of I2 = 2 A.  From the value of I2 the value of i1 = [30 – (3 x I2 )] / 8 = 3 A and I3 = I2 /2 =1 A.  V1 = 8 x I1 = 24 V,  V2 = 3 x I2 = 6 V,  V3 = 6 x I3 = 6 V. 57 29 30 31 32 33