EE2301: Basic Electronic Circuit Quick Summary of Last Lecture Block A Unit 1 Three Basic Laws EE2301: Block A Unit 21

Block A Unit 12 Fundamental law for charge  Current has to flow in closed loop  No current flows if there is a break in the path  Underlying physical law: Charge cannot be created or destroyed  This is the basis of Kirchhoff’s Current Law i1i1 i4i4 i3i3 i2i2 Kirchhoff’s current law Sum of currents at a node must equal to zero: i 1 + i 2 + i 3 + i 4 = 0

Block A Unit 13 Fundamental law on voltage  Energy is required to push electrons through a resistive element  That same energy needs to be generated by a source  Total energy generated in a circuit must equal total energy consumed in the circuit  Energy cannot be created or destroyed  Therefore, voltage rise = voltage drop - V V 1 - -V4+-V4+ +V2-+V2- Kirchhoff’s voltage law Net voltage around a closed circuit is zero: v 1 + v 2 + v 3 + v 4 = 0

Block A Unit 14 Resistance and Ohm’s Law i v V I 1/R Ohm’s law: V = IR Ideal RESISTOR shows linear resistance obeying Ohm’s law When current flows through any circuit element, there will always be a resistance to its flow which results in a voltage drop across that circuit element + _ IMPORTANT: Positive current is defined here as flowing from higher to lower voltage (Remember) Unit: Ohm (Ω) A L ρ: resistivity (material property) A: cross-sectional area

Block A Unit 15 Parallel network (Highlights) R1R1 R2R2 IsIs RNRN I1I1 I2I2 ININ RPRP IsIs Equivalent Resistance 1/R P = 1/R 1 + 1/R 2 + …+ 1/R N Current divider rule

Block A Unit 16 Series network (Highlights) R1R1 R2R2 VsVs +-+- RSRS VsVs +-+- Equivalent Resistance R S = R 1 + R 2 + …+ R N RNRN Voltage divider rule V N = V S (R N /R S ) +-+- V1V V2V VNVN

Let’s have a look first EE2301: Block A Unit 27

8 Block A Unit 2 outline  Applying the 3 laws to analyze DC circuits  Systematic methods for analysis > Nodal voltage analysis (application of KCL and Ohm’s law) > Mesh current analysis (application of KVL and Ohm’s law) > Superposition (can be a powerful tool) G. Rizzoni, “Fundamental of EE” Chapter 3.2 – 3.5

EE2301: Block A Unit 29 Nodal voltage analysis V1V1 X R2R2 R3R3 R1R1 V2V2 V3V3 Nodal voltage analysis is simply an application of Ohm’s law and KCL together. Here we express branch currents in terms of voltage and resistance using Ohm’s law Applying NVA at node X: This can be seen if we first consider currents in each branch arriving at X: Current from V 1 to V X via R 1 = (V 1 - V X )/R 1 Current from V 2 to V X via R 2 = (V 2 - V X )/R 2 Current from V 3 to V X via R 3 = (V 3 - V X )/R 3 Summing these together, we obtain the above final expression

EE2301: Block A Unit 210 NVA example 1 Problem 3.1: Use nodal voltage analysis to find the voltages V 1 and V 2 1 Ω Apply KVL at V 1 :Apply KVL at V 2 : Solve for V 1 and V 2 : V 1 = 4.8 V, V 2 = 2.4 V

EE2301: Block A Unit 211 NVA example 2 Problem 3.4: Use nodal voltage analysis to find the current through the voltage source First, define and label the unknown nodes Apply KCL at V 1 : Apply KCL at V 2 : Apply KCL at V 3 :

EE2301: Block A Unit 212 NVA example 2 solution This slide is meant to be blank There are now 3 equations but 4 unknowns; we need one more equation!! V 3 - V 2 = 3 Now, eliminate V 3 from all 3 equations: V 2 + (3 + V 2 ) - 2V 1 = 1  V 1 - V 2 = 1  Now, eliminate V 2 :

EE2301: Block A Unit 213 Mesh current analysis vsvs + _ R1R1 R2R2 R3R3 R4R4 0V i1i1 i2i2 Mesh current analysis is simply an application of Ohm’s law and KVL together. Here we express branch voltages in terms of current and resistance using Ohm’s law Around Mesh 1: v s = i 1 (R 1 + R 2 ) - i 2 R 2 Around Mesh 2: i 2 (R 3 + R 4 + R 2 ) - i 1 R 2 = 0 Apply KVL to each mesh in turn  If the current direction is known, then the easiest choice is simply to follow it. This will avoid any confusion (e.g. in a voltage source, define the current as flowing in the direction of voltage gain)  Pay close attention to the direction of one mesh current to another (e.g. in this instance, i 1 is flowing opposite to i 2 hence we take the difference in R 2

EE2301: Block A Unit 214 MCA example 1 This slide is meant to be blank Problem 3.30 Use mesh current analysis to find the current (i) through the 1/5 Ω resistor

EE2301: Block A Unit 215 MCA example 1 solution This slide is meant to be blank Since I = i 1, so only 2 unknown meshes to solve for KVL around mesh 2:KVL around mesh 3: Solving for I 2 and I 3 : I 2 = 20/31 A, I 3 = 15/31 A Current through the 1/5 Ω resistor, i = i3 - i2 = - 5/31 = A

EE2301: Block A Unit 216 MCA example 2 Problem 3.17 Use mesh current analysis to find the voltage across the current source I3I3 Apply KVL around mesh 1: 2 = I 1 (2+3) - I 2 (3)  5I 1 - 3I 2 = 2 Apply KVL around mesh 2: -V = I 2 (3+1) - I 1 (3)  4I 2 - 3I 1 = -V Apply KVL around mesh 3: V = I 3 (3+2)  5I 3 = V

EE2301: Block A Unit 217 MCA example 2 This slide is meant to be blank There are now 3 equations but 4 unknowns; we need one more equation!! 2 = I 3 - I 2 Substitute into mesh 3 to eliminate i 3 and use this to eliminated i 2 in mesh 2: Substitute into mesh 1 to find V:

EE2301: Block A Unit 218 Principle of superposition AB Consider what happens when you throw a stone into a pool at A & B A B Both A and B together Only at B B A Only at A If the stones hit A and B together, the result will be a combination of the individual responses of A and B. This is the principle of superposition.

EE2301: Block A Unit 219 Superposition in circuits B A B’ A’ INPUTOUTPUT SYSTEM A+B A’+B’ + - VGVG IBIB RBRB RGRG I Say we want to find the current through R B If we apply superposition, this current is the sum of the individual currents corresponding to each of the sources in the circuit, i.e. current associated only with V G or I B alone. That is to say, we need to remove the effects from all the sources except one, and find the corresponding value of I for this source. We then repeat this for the other sources. But how do we remove the effects of a certain source?

EE2301: Block A Unit 220 Disabling sources in superposition Voltage source: We want no voltage drop across Therefore replace with a short Current source: We want no current flowing through Therefore replace with an open circuit IBIB RBRB RGRG I1I1 + - VGVG RBRB RGRG I2I2 V G = 0 I B = 0 I = I 1 + I 2 Open circuit Short circuit VsVs +-+- ININ

EE2301: Block A Unit 221 Superposition in Circuits + - VGVG IBIB RBRB RGRG I IBIB RBRB RGRG I1I1 + - VGVG RBRB RGRG I2I2 V G = 0I B = 0 I = I 1 + I 2 Open circuit Short circuit Find I

EE2301: Block A Unit 222 Superposition example 1 Problem 3.40 Determine, using superposition, the current through R 1 due only to the source V S2

EE2301: Block A Unit 223 Superposition example 1 solution This slide is meant to be blank Re-drawn circuit due to V S2 only Note that now R 1 || R 2 are parallel Suggested strategy: 1)Use voltage divider rule to find voltage across R 1 2)Use ohm’s law to find current through R 1

EE2301: Block A Unit 224 Superposition example 2 This slide is meant to be blank Problem 3.41 Determine, using superposition, the voltage across R

EE2301: Block A Unit 225 Superposition example 2 This slide is meant to be blank V R due to V G only: R and R B are parallel, which together are in series with R G Apply voltage divider rule: V R due to I B only: All 3 resistor are in parallel Adding the two solutions together: V R = 5.99V