M. Afendi M. Piah, PhD Circuit Theory y (SKEE 1023) Assoc. Prof. Dr. Mohamed Afendi Mohamed Piah Institute of High Voltage & High Current Faculty of Electrical Engineering, UTM 1 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Topics Methods M th d off Circuit Ci it Analysis: A l i Nodal N d l Analysis; Node Voltages; Mesh A l i Analysis; L Loop C Currents; t Ci it Circuit Analysis with Voltage and Current I d Independent d t and d D Dependent d t Sources. 2 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory METHODS OF CIRCUIT ANALYSIS Introduction Two powerful techniques for circuit analysis, ie; ¾ Nodal analysis : based on application of KCL. ¾ Mesh analysis : based on application of KVL. Analyze ay e a any y linear ea ccircuit cu t by obta obtaining g a set o of ssimultaneous u ta eous equat equations o s that are then solved to obtain the required values of current or voltage. Nodal Analysis Provide a general procedure for analyzing circuits using node voltages as the circuit variables. Interested in finding the node voltages. voltages 3 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Steps p to Determine Node Voltages g ((Nodal Analysis) y ) 1. Select a node as the reference node. Assign voltages v1, v2, …. Vn-1 to the remaining n – 1 nodes. 2 Apply KCL to each of the n – 1 nonreference nodes. 2. nodes 3. Solve the resulting simultaneous equations to obtain the unknown node voltages. The reference (datum) node is commonly called the ground (assumed to have zero potential). Symbols : -Common ground Ground -Ground -Chassis ground 4 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Method of Nodal Analysis 5 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory At node d 1 and d node d 2, 2 apply l KCL gives; i I1 = I 2 + i1 + i2 and I 2 + i2 = i3 v1 − 0 i1 = = G1v1 R1 v1 − v2 = G2 (v1 − v2 ) i2 = R2 i3 = v2 − 0 = G3v2 R3 6 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory At node 1 ⎛1 1 ⎞ 1 ⎜ ⎟ I1 = I 2 + ⎜ + v1 − v2 R2 ⎝ R1 R2 ⎠ ⎛1 1 ⎞ 1 ⎜⎜ + ⎟⎟v1 − v2 = I1 − I 2 R2 ⎝ R1 R2 ⎠ or (G1 + G2 )v1 − G2v2 = I1 − I 2 ⎛ 1 1 1 ⎞ ⎜ I 2 = − v1 + ⎜ + ⎟⎟v2 R2 ⎝ R2 R3 ⎠ At node 2 − ⎛ 1 1 1 ⎞ v1 + ⎜⎜ + ⎟⎟v2 = I 2 R2 ⎝ R2 R3 ⎠ or − G2 v1 + (G2 + G3 )v2 = I 2 INSPIRING CREATIVE AND INNOVATIVE MINDS 7 M. Afendi M. Piah, PhD Circuit Theory In matrix form; 1 ⎡1 + ⎢R R 2 ⎢ 1 ⎢ − 1 ⎢⎣ R2 or 1 ⎤ R2 ⎥ ⎡ v1 ⎤ ⎡ I1 − I 2 ⎤ ⎥ = 1 1 ⎥ ⎢⎣v2 ⎥⎦ ⎢⎣ I 2 ⎥⎦ + R2 R3 ⎥⎦ ⎡G1 + G2 ⎢ −G 2 ⎣ − G2 ⎤ ⎡ v1 ⎤ ⎡ I1 − I 2 ⎤ =⎢ ⎥ ⎢ ⎥ G2 + G3 ⎦ ⎣v2 ⎦ ⎣ I 2 ⎥⎦ − Then, determine the values of v1 and v2. 8 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Example 2.1 Calculate the node voltages in the circuit. 9 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Problem 2.1 Obtain the node voltages in the circuit. Ans: v1 = -2 V V, v2 = -14 V 10 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Example 2 2.2 2 Determine the voltages at the nodes. 11 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Problem 2.2 22 Determine the voltages at the nodes. Ans : v1 = 80 V, v2 = -64 V, v3 = 156 V 12 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Nodal Analysis with Voltage Sources How voltage H lt sources affect ff t nodal d l analysis. l i Consider C id the th following f ll i t two possibilities. Case 1: ¾ If a voltage source is connected between the reference node and a nonreference f node; d ⇒ v1 = 10 V. V Case 2: ¾ If the voltage source (dependent or i d independent) d t) is i connected t d between b t two nonreference nodes. The nonreference nodes form a generalized node or supernode. 13 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory ¾ Nodes 2 and 3 form a supernode. ¾ There is no way knowing the current through a voltage source. source ¾ KCL must be satisfied at a supernode, i.e. i1 + i4 = i2 + i3 or v1 − v2 v1 − v3 v2 v3 + = + 2 4 8 6 and v2 − v3 = 5 14 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Example 2.3 For the circuit shown below, find the node voltages. Applying A l i KCL to the h supernode; d 2 = i1 + i2 + 7 where, v v i1 = 1 ; i2 = 2 and v2 − v1 = 2 2 4 INSPIRING CREATIVE AND INNOVATIVE MINDS 15 M. Afendi M. Piah, PhD Circuit Theory Problem 2.3 Find v and i. Ans: -0.2 V, 1.4 A 16 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Example 2.4 Find the node voltages in the circuit. Solution KCL at supernode (1 - 2) 10 + i3 = i1 + i2 where, i3 = v3 − v2 v −v v ; i1 = 1 4 ; i2 = 1 6 3 2 KCL at supernode (3 - 4) i1 = i3 + i4 + i5 where, i1 , and i3 same as above and; i4 = v v4 ; i5 = 3 1 4 Supportive equations : v1 − v2 = 20 and v3 − v4 = 3v x = 3(v1 − v4 ) 17 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Problem 2.4 24 Find v1, v2 and v3 in the circuit using nodal analysis. Ans: v1 = 3.043 V, v2 = -6.956 V, v3 = 0.6522 18 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Mesh Analysis Another general procedure for analyzing circuits, using mesh current as the circuit variables. Definations: Loop is a closed path with no node passed more than once. once Mesh is a loop that does not contain any other loop within it. Mesh analysis applies KVL to find unknown currents. Only applicable to a circuit that is planar (two-dimensional). Nonplanar circuits can be handled using nodal analysis. 19 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Types of circuit A planar circuit The same ccircuit cu t redrawn ed a A nonplanar circuit 20 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory ¾ Paths abefa and bcdeb are meshes. i1 and i2 are mesh current. ¾ In mesh analysis, we are interested in applying KVL to find the mesh current. Steps to Determine Mesh Currents Assign mesh currents i1, i2, …. in to the n meshes. Assume mesh currents flow clockwise. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh current. Solve the resulting n simultaneous equations to get the mesh currents. 21 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory For mesh 1 − V1 + R1i1 + R3 (i1 − i2 ) = 0 or (R1 + R3 )i1 − R3i2 = V1 For mesh 2 R2i2 + V2 + R3 (i2 − i1 ) = 0 In matrix form; ⎡ R1 + R3 ⎢ −R 3 ⎣ ¾ − R3 ⎤ ⎡ i1 ⎤ ⎡ V1 ⎤ = R2 + R3 ⎦⎥ ⎢⎣i2 ⎥⎦ ⎣⎢− V2 ⎥⎦ or − R3i1 + (R2 + R3 )i2 = −V2 Notice that the branch currents are different from the mesh currents unless the mesh is isolated. i for a mesh current and I for a branch current. I1 = i1 ; I 2 = i2 ; I 3 = i1 − i2 INSPIRING CREATIVE AND INNOVATIVE MINDS 22 M. Afendi M. Piah, PhD Circuit Theory Example 2.5 Find the branch currents I1, I2 and I3 using mesh analysis. Solution Apply KVL for mesh 1 − 15 + 5i1 + 10(i1 − i2 ) + 10 = 0 or 3i1 − 2i2 = 1 Apply KVL for mesh 2 6i2 + 4i2 + 10(i2 − i1 ) − 10 = 0 In matrix form; ⎡ 3 − 2⎤ ⎡ i1 ⎤ ⎡1⎤ ⎢− 1 2 ⎥ ⎢i ⎥ = ⎢1⎥ ⎣ ⎦⎣ 2 ⎦ ⎣ ⎦ Ans : i1 = 1 A ; i2 = 1 A or − i1 + 2i2 = 1 Therefore, the branch currents are; I1 = i1 = 1 A; I 2 = i2 = 1 A; I 3 = i1 − i2 = 0 A INSPIRING CREATIVE AND INNOVATIVE MINDS 23 M. Afendi M. Piah, PhD Circuit Theory Problem 2.5 Calculate the mesh current i1 and i2. Ans: i1 = 2/3 / A, i2 = 0 A 24 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Example 2.6 Use mesh analysis to find the current Io in the circuit. Solution Apply KVL for mesh 1 − 24 + 10(i1 − i2 ) + 12(i1 − i3 ) = 0 or 11i1 − 5i2 − 6i3 = 12 Apply KVL for mesh 2 24i2 + 4(i2 − i3 ) + 10(i2 − i1 ) = 0 Apply KVL for mesh 3 4 I 0 + 12(i3 − i1 ) + 4(i3 − i2 ) = 0 or − 5i1 + 19i2 − 2i3 = 0 and I 0 = i1 − i2 ; so that 4(i1 − i2 ) + 12(i3 − i1 ) + 4(i3 − i2 ) = 0 Ans : I 0 = i1 − i2 = 2.25 − 0.75 = 1.5 A or − i1 − i2 + 2i3 = 0 INSPIRING CREATIVE AND INNOVATIVE MINDS 25 M. Afendi M. Piah, PhD Circuit Theory P bl Problem 26 2.6 Using mesh analysis, find I0 in the circuit. Ans: -5 A 26 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Mesh Analysis y with Current Sources The presence of the current sources reduces the number of equations. The number of equations (based on KVL) that need to be write is n – m, where n: number of meshes and m: number of current sources. Consider the following 2 possible cases: Case 1: When current source exists only in one mesh. In the figure, n − m = 2 − 1 = 1 ; therefore, KVL at mesh 1: − 10 + 4i1 + 6(i1 − i2 ) = 0 and d i2 = −5 (supportiv ( ti e equation) ti ) 27 INSPIRING CREATIVE AND INNOVATIVE MINDS M. Afendi M. Piah, PhD Circuit Theory Case 1: When current source exists between two meshes. We create a supermesh by excluding the current source and any elements connected in series i with i h it. i supermesh KVL at supermesh gives : − 20 + 6i1 + 10i2 + 4i2 = 0 or 6i1 + 14i2 = 20 Supportive equation : i2 − i1 = 6 ⇒ (a) ⇒ (b) Solving eqns (a) and (b), we get, i1 = −3.2A ; i2 = 2.8 A INSPIRING CREATIVE AND INNOVATIVE MINDS 28 M. Afendi M. Piah, PhD Circuit Theory Example 2.7 Find i1 to i4 using mesh analysis in the circuit. Solution Number of meshes, n = 4 ; Number of Current Sources, m = 2 ⇒ Number of equations need to be applied, n − m = 4 − 2 = 2 Meshes 1 & 2 form a supermesh. Also meshes 2 & 3 form another supermesh. p The two supermeshes intersect and form a larger supermesh. ⇒ Applying KVL to the larger supermesh; 2i1 + 4i3 + 8(i3 − i4 ) + 6i2 = 0 Supportive equations; ⇒ (c) or i1 + 3i2 + 6i3 − 4i4 = 0 ⇒ i2 − i1 = 5 ⇒ i2 − i3 = 3I 0 = 3(−i4 ) ⇒ (d) ⇒ (a) ⇒ Applying KVL in mesh 4; 2i4 + 8(i4 − i3 ) + 10 = 0 or 5i4 − 4i3 = −5 ⇒ (b) From eqns (a) to (d), we get, i1 = −7.5 A, i2 = −2.5 A, i3 = 3.93 A, i4 = 2.143 A INSPIRING CREATIVE AND INNOVATIVE MINDS 29 M. Afendi M. Piah, PhD Circuit Theory Problem 2.7 27 Use mesh analysis to determine i1, i2 and i3. Ans: i1 = 3.474 A; i2 = 0.4737 A; i3 = 1.1052 A 30 INSPIRING CREATIVE AND INNOVATIVE MINDS