LECTURE # 3 & 4 Circuit Laws Network Analysis Kirchhoff’s Current Law Kirchhoff’s Voltage Law Superposition Theorem Thevenin’s Theorem Norton’s Theorem 𝑺𝒆𝒍𝒇 𝑺𝒕𝒖𝒅𝒚 Common Electrical Circuit and Network Terms n/n Term 1 Electric Circuit 2 Circuit Param eters 3 Node 4 Branch 5 Lum ped Param eters 6 Distributed Param eters 7 Loop 8 Linear Circuit 9 Non-Linear Circuit 10 Bilateral Circuit 11 Unilateral Circuit 12 Short Circuit 13 Electric Network 14 Passive Network 15 Acive network 16 M esh 17 Open Circuit 18 Constant Current Source 19 Equivalent Resistance 20 Constant Voltage Source Description KIRCHHOFF’S LAWS Kirchhoff’s Current Law – [KCL] The sum of the currents entering 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4 = 𝐼𝑎 + 𝐼𝑐 + 𝐼𝑥 a node is always equal to the sum 𝐼 + 𝐼 + 𝐼 + 𝐼 + (−𝐼 − 𝐼 − 𝐼 ) = 0 𝑎 𝑐 𝑥 of the currents leaving that node. 1 2 3 𝑛4 𝐼𝑘 = 0 𝑰𝟐 𝑘=1 𝑰𝒙 𝑰𝒂 Example #: 𝑰𝟒 𝑰𝟏 𝑰𝟑 𝑰𝒄 The algebraic sum of the currents at a node is always equal to zero. Those currents entering considered positive while leaving are negative 𝐼1 𝐼2 𝐼5 𝐼6 𝐼4 𝐼3 𝑰𝟏 = 𝟐𝑨 𝑰𝟓 =? ? 𝑰𝟓 = 𝟒𝑨 𝑰𝟐 = 𝟐𝑨 𝑰𝟒 = 𝟐𝑨 𝑰𝟑 = 𝟐𝑨 𝑰𝟔 =? ? 𝑰𝟔 = 𝟎𝟎𝟎𝟎𝟎𝑨 Kirchhoff’s Voltage Law – [KVL] According to Kirchhoff’s Voltage Law the sum of the emf (voltage) sources acting upon a loop is equal to the sum of the voltage drops experienced along the loop. 𝑛 𝑛 𝐸𝑘 = 𝑘=1 𝑹𝟐 𝐼𝑅𝑘 𝑘=1 𝑹𝟑 𝑹𝟏 𝑬𝟐 𝐸1 + (−𝐸2 ) = 𝐼𝑅1 + 𝐼𝑅2 + 𝐼𝑅3 + 𝐼𝑅4 𝑬𝟏 𝑹𝟒 𝐸1 + −𝐸2 − 𝐼 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 = 0 The direction of current is initially assumed, since the resultant emf is not known Resultant magnitude and polarity of sources will determine the larger source 𝑰 Example #: 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝐶𝑢𝑟𝑟𝑒𝑛𝑡𝑠 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑎𝑙𝑙 𝑏𝑟𝑎𝑛𝑐ℎ𝑒𝑠 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 ℎ𝑒𝑙𝑝 𝑜𝑓 𝐾𝐶𝐿 𝑎𝑛𝑑 𝐾𝑉𝐿 b c 𝑰𝒂 1. Identify currents in the 4V 1Ω branches 0.5Ω 2. Choose any closed loops 3. Assume the direction of d a 3Ω current flow 2V 𝑰𝒃 4. Equate the sum of voltage drops to emfs 3V 𝑰 𝒄 5. Equate the algebraic sums of e 1Ω f emfs and voltage drops to Consider loop abcda zero 1 ∗ 𝐼𝑎 + 0.5𝐼𝑎 + 3𝐼𝑏 = 2 + 4 6. Find currents at a node 1.5𝐼𝑎 + 3𝐼𝑏 = 6 𝑨𝒄𝒄𝒐𝒓𝒅𝒊𝒏𝒈 𝒕𝒐 𝑲𝑪𝑳: 7. Solve simultaneous equations 𝑰𝒂 + 𝟐𝑰𝒃 = 𝟒 For loop aefda 1 ∗ 𝐼𝑐 − 3𝐼𝑏 = 3 − 2 𝐼𝑐 − 3𝐼𝑏 = 1 −𝒊 𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒚 𝒊 𝒃𝒚 𝟑 𝟑𝑰𝒄 − 𝟗𝑰𝒃 = 𝟑 −𝒊𝒊𝒊 𝑰𝒂 = 𝑰𝒃 + 𝑰𝒄 −𝒊𝒊 𝑰𝒄 + 𝟑𝑰𝒃 = 𝟒 −𝒊𝒗 𝑨𝒅𝒅 𝒊𝒊𝒊 𝒂𝒏𝒅 𝒗 𝑰𝒂 = 𝟐. 𝟓𝑨𝑰𝒃 + 𝑰𝒄 𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒚 𝒊𝒗 𝒃𝒚 𝟑 𝟑𝑰𝒄 + 𝟗𝑰𝒃 = 𝟏𝟐 −𝒗 Open Circuit An open circuit is a circuit where the path has been interrupted or "opened" at some point so that current will not flow Active Circuit [with sources and circuit elements] Open-circuit voltage (OCV or 𝑉𝑜𝑐 ) 𝒂 is the difference of electrical potential between two terminals of a 𝑉𝑜𝑐 device when disconnected from any 𝑉𝑎𝑏 circuit 𝒃 The open circuit condition is the same as a circuit with infinite resistance No current flows through an open circuit P.D. measured between terminals of an open circuit is the same as source voltage for a simple d.c. circuit Short Circuit 𝑺𝒆𝒍𝒇 𝑺𝒕𝒖𝒅𝒚 NETWORK ANNALYSIS R1 LECTURE # 6 E1 – Voltage Source R2 Superposition Theorem E2 – Voltage Source R3 A-? R4 IA – Current Source Superposition Theorem E1 R3 In any linear network with more than one source (voltage source/current source) the resultant current flowing I4 through any branch is the sum of the individual currents due to the different sources considered separately, whereby (voltage sources are R1 R4 R 2 replaced by short circuits or their internal resistances and current sources by an open circuit or E2 infinite resistance) The direction of the resultant currents are determined by the size of the and direction of values of currents flowing 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝐼4 𝑖𝑠 𝑎 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑡ℎ𝑒 along the branch. 𝑒𝑓𝑓𝑒𝑐𝑡 𝑜𝑓 𝑠𝑜𝑟𝑐𝑒𝑠 𝑬𝟏 . 𝑬𝟐 𝑎𝑛𝑑 𝑰𝑨 IA Current due to Voltage Source E1 𝑰𝑬𝟏 − 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒖𝒆 𝒕𝒐 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟏 𝑬𝟏 𝐈𝐄𝟏 = 𝑹𝑬𝟏 E1 R3 𝐼𝐸1 R1 E1 R3 𝐼𝐸1 R1 𝐼4𝐸1 R2 R4 𝐼4𝐸1 R4 R2 𝑂𝑝𝑒𝑛 𝐶𝑖𝑟𝑐𝑢𝑖𝑡 𝑆ℎ𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡 𝑹𝑬𝟏 − 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒄𝒊𝒓𝒄𝒖𝒊𝒕 𝑹𝑬𝟏 = 𝑹𝟏 + 𝑹𝟐 ∗ (𝑹𝟑 + 𝑹𝟒) 𝑹𝟐 + 𝑹𝟑 + 𝑹𝟒 𝑰𝟒𝑬𝟏 − 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒓𝒆𝒔𝒊𝒔𝒕𝒐𝒓 𝑹𝟒 𝒅𝒖𝒆 𝒕𝒐 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟏 𝐈𝐄𝟏 ∗ 𝑹𝟐 𝑰𝟒𝑬𝟏 = 𝑹𝟐 + 𝑹𝟑 + 𝑹𝟒 Current due to Voltage Source E2 𝑰𝑬𝟐 − 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒖𝒆 𝒕𝒐 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟐 𝑬𝟐 𝐈𝐄𝟐 = 𝑹𝑬𝟐 𝑆ℎ𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡 R3 𝐼4𝐸2 𝐼𝐸2 R1 R2 R4 E2 𝑹𝑬𝟏 − 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒄𝒊𝒓𝒄𝒖𝒊𝒕 𝑹𝟏 ∗ (𝑹𝟑 + 𝑹𝟒) 𝑹𝑬𝟐 = 𝑹𝟐 + 𝑹𝟏 + 𝑹𝟑 + 𝑹𝟒 𝑰𝟒𝑬𝟐 − 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒓𝒆𝒔𝒊𝒔𝒕𝒐𝒓 𝑹𝟒 𝒅𝒖𝒆 𝒕𝒐 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟐 𝑰𝟒𝑬𝟐 = 𝐈𝐄𝟐 ∗ 𝑹𝟐 𝑹𝟐 + 𝑹𝟑 + 𝑹𝟒 Current due to Voltage Source IA 𝑆ℎ𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡 R3 𝑰𝟒𝑨 R4 R1 𝑰𝑨 R2 𝑆ℎ𝑜𝑟𝑡 𝐶𝑖𝑟𝑐𝑢𝑖𝑡 𝑰𝟒𝑨 − 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒓𝒆𝒔𝒊𝒔𝒕𝒐𝒓 𝑹𝟒 𝒅𝒖𝒆 𝒕𝒐 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒔𝒐𝒖𝒓𝒄𝒆𝑰𝑨 𝑶𝒏𝒍𝒚 𝒂𝒑𝒑𝒍𝒚 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒆𝒗𝒊𝒅𝒆𝒓 𝒓𝒖𝒍𝒆 𝒕𝒐 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒃𝒓𝒂𝒏𝒄𝒉 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝑰𝟒𝑨 𝑹 𝑹𝟐 [ 𝑹 𝟏+ 𝑹 + 𝑹𝟑] 𝟏 𝟐 𝑰𝟒𝑨 = 𝑰𝑨 ∗ 𝑹 𝑹𝟐 [ 𝑹 𝟏+ 𝑹 + 𝑹 𝟑] + 𝑹 𝟒 𝟏 𝟐 Resultant Current Due to All Sources 𝑰𝟒𝑬𝟏 𝑰𝟒 𝑰𝟒𝑨 𝑰𝟒𝑬𝟐 𝑹𝒆𝒔𝒖𝒍𝒕𝒂𝒏𝒕 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒅𝒖𝒆 𝒕𝒐 𝒕𝒉𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕 𝒔𝒐𝒖𝒓𝒄𝒆𝒔 𝑰𝟒 𝒊𝒔 𝒕𝒉𝒆 𝒔𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕 𝒄𝒖𝒓𝒓𝒆𝒏𝒕𝒔 𝒅𝒖𝒆 𝒕𝒐 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕 𝒔𝒐𝒖𝒓𝒄𝒆𝒔 𝑰𝟒 = 𝑰𝟒𝑬𝟏 + 𝑰𝟒𝑬𝟐 + 𝑰𝟒𝑨 Example#: 𝐸1 = 12𝑉 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝐸2 = 24𝑉 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼𝐴 = 𝐴4 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑅1 = 4Ω 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟 𝑅2 = 2Ω 𝑅2 =? ? 𝑅3 = 6Ω 𝑅4 = 4Ω R3 E1 I2 R1 R2 R4 E2 IA EEB 211/231 & GEC 255 CONTACT TIME TABLE EEB 211/LAB/01/2 Monday 08:00 AM 10:00 AM BLDG 251 ROOM 201 Lab Work Group 1 &2 EEB 211/231/Lec Monday 04:00 PM 05:30 PM BLDG 252 ROOM R4 Lecture [26 August] EEB 211/231/Lec Tuesday 04:00 PM 05:30 PM BLDG 252 ROOM R4 Lecture [27 august] EEB 211/LAB/3/4 Wednesday 08:00 AM 10:00 AM BLDG 251 ROOM 201 Lab Work Group 2/3 EEB 211/LAB/5/6 Thursday 08:00 AM 10:00 AM BLDG 251 ROOM 201 Lab Work Group 5/6 EEB 211/LAB/7/8 Monday 02:00 PM 04:00 PM BLDG 251 ROOM 201 Lab Work Group 7/8 EEB 211/LAB/9/10 Wednesday 03:00 PM 05:00 PM BLDG 251 ROOM 201 Lab Work Group 9/10 GEC 255 Thursday 12:00 AM 13:00 PM BLDG 248 ROOM 011 Tutorials EEB 211/231/Lec Thursday 02:00 PM 04:00 PM BLDG 252 ROOM G1 Lecture EEB 211/TUT/01 Monday 11:00 AM 12:00 PM BLDG 248 ROOM 011 Tutorial Group 1 EEB 211/TUT/02 Tuesday 11:00 AM 12:00 PM BLDG 248 ROOM 009 Tutorial Group 2 EEB 211/TUT/03 Thursday 11:00 AM 12:00 PM BLDG 248 ROOM 009 Tutorial Group 3 GEC 255 Friday 11:00 AM 13:00 PM BLDG 248 ROOM 009 Lecture EEB 211/TUT/04 Friday 11:00 AM 12:00 PM BLDG 248 ROOM 009 Tutorial Group 4 Example # R3 𝑷𝟑 =? E2 𝑰𝟐 =? R1 R4 R2 IA E1 𝑹𝟏 = 𝑹𝟒 = 𝟐Ω 𝑰𝑨 = 𝟖𝑨 𝑹𝟐 = 𝟒Ω, 𝑹𝟑 = 𝟑Ω 𝑬𝟏 = 𝟏𝟎𝑽 𝑬𝟐 = 𝟔𝑽 Solution 𝑷𝟑 =? 𝑰𝟐 = 𝑰𝟐𝑬𝟏 + 𝑰𝟐𝑬𝟐 + 𝑰𝟐𝑨 𝑰𝟐 =? 𝑷𝟑 = 𝑰𝟐 𝟑 ∗ 𝑹𝟑 𝑰𝟑 = 𝑰𝟑𝑬𝟏 + 𝑰𝟑𝑬𝟐 + 𝑰𝟑𝑨 𝑼𝒔𝒊𝒏𝒈 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟏 =? 𝐼𝐸1 = R3 R1 R2 R4 E1 𝑰𝟐𝑬𝟏 R3 R2 E1 R04 𝑹𝟎𝟒 = 𝟎 ∗ 𝑹𝟒 𝟎 + 𝑹𝟒 𝑹𝑬𝟏 = 𝑹𝟐 + 𝑰𝟐𝑬𝟏 = 𝑰𝑬𝟏 R0 𝐼3𝐸1 = 𝐼𝐸1 ∗ 𝑰𝟑𝑬𝟏 R1 𝐸1 𝑅𝐸1 𝑅1 𝑅1 + 𝑅3 𝑹𝟏 ∗ 𝑹𝟐 𝑹𝟏 + 𝑹𝟐 Solution 𝑼𝒔𝒊𝒏𝒈 𝒔𝒐𝒖𝒓𝒄𝒆 𝑰𝑨 =? 𝑅21 ∗ R3 R1 𝑅31 ∗ R2 R4 IA R0 R3 𝑰𝟐𝑨 R2 R0 IA R04 𝑹𝟎𝟒 = 𝑅3 ∗ 𝑅1 𝑅3 + 𝑅1 R0 𝐼3𝐴 𝑅21 = 𝐼𝐴 ∗ 𝑅21 + 𝑅3 𝐼2𝐴 𝑅31 = 𝐼𝐴 ∗ 𝑅31 + 𝑅2 𝑰𝟑𝑨 R1 𝑅2 ∗ 𝑅1 𝑅2 + 𝑅1 𝟎 ∗ 𝑹𝟒 = 𝟎Ω 𝟎 + 𝑹𝟒 Solution 𝑼𝒔𝒊𝒏𝒈 𝒔𝒐𝒖𝒓𝒄𝒆 𝑬𝟐 =? 𝑰 𝟑𝑬𝟐 R3 𝑰𝟐𝑬𝟐 R1 𝑰𝑬𝟐 𝑬𝟐 = 𝑹𝑬𝟐 𝑰𝑬𝟐 E2 R4 R2 𝑹 = ∞ 𝑨 𝑹𝟑𝟐𝟏 = 𝑹𝟑 + 𝑰𝟑𝑬𝟐 R3 R0 𝑹𝟐 ∗ 𝑹𝟏 𝑹𝟐 + 𝑹𝟏 𝑰𝟑𝑬𝟐 𝑰𝟐𝑬𝟐 𝑹𝟏 = 𝑰𝟑𝑬𝟐 ∗ 𝑹𝟏 + 𝑹𝟐 𝑰𝟐 = 𝑰𝟐𝑬𝟏 + −𝑰𝟐𝑬𝟐 + (−𝑰𝟐𝑨 ) 𝑰𝑬𝟐 𝑰𝟐𝑬𝟐 R1 𝑹𝟒 ∗ 𝑹𝟑𝟐𝟏 𝑹𝟒 + 𝑹𝟑𝟐𝟏 𝑹𝟒 = 𝑰𝑬𝟐 ∗ 𝑹𝟒 + 𝑹𝟑𝟐𝟏 R0 R2 𝑹𝑬𝟐 = R4 E2 𝑰𝟐 = 𝑰𝟐𝑬𝟏 − 𝑰𝟐𝑬𝟐 − 𝑰𝟐𝑨 𝑰𝟑 = 𝑰𝟑𝑬𝟏 + (−𝑰𝟑𝑬𝟐 ) + 𝑰𝟑𝑨 𝑰𝟑 = 𝑰𝟑𝑬𝟏 − 𝑰𝟑𝑬𝟐 + 𝑰𝟑𝑨 𝑷𝟑 = 𝑰𝟐 𝟑 ∗ 𝑹𝟑