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Download Helpful for engineering students and more Slides Science education in PDF only on Docsity! ELECTRICAL CIRCUITS Mesh-current analysis (Maxwell's Theorem) AND Nodal analysis Mesh-current and Nodal analysis may be applied to both d.c. and a.c. circuits. C Kqwerawera 07/26/2021 ELECTRICAL CIRCUITS Mesh-current analysis Mesh-current analysis is merely an extension of the use of Kirchhoff's laws. The figure shows a network whose circulating mesh currents or loop currents I, I, and I, have been assigned to closed loops in the circuit rather than to branches. 07/26/2021 ELECTRICAL CIRCUITS Use mesh-current analysis to determine the current flowing in (a) the 5 Q, resistance, and (6) the 1 Q resistance of the d.c. circuit 07/26/2021 ELECTRICAL CIRCUITS Using the mesh currents I,, 1, and I, shown and Kirchhoff 's voltage law: (3 +5), —3lo=—4 (4414645)lo— (5), — (Ds = 0 (148) — (lp = —5 81, — 5I> —4=0 — Sf, + 16/2 —TIs3 =O 07/26/2021 ELECTRICAL CIRCUITS Using determinants qy —t2 Ts —5 oO —4/ | 8 o —4|/ |s8 —5 —4 16 —1 O —5 -—1 0 —5 16 O —1 9 5 oO 9 5 Oo -1 5 —1 “Ts —5 O —5 16 —1 o -1 9 I _ Is —1 0 16. —1l); ~|;—1 O —5 —I sl 5|-4 | 5 sl-4lo 9 | 9 5 -1 9 9 5 o 9 -a] ie] | 8 Ve oO -1 —5 16 —1 ~ "716 —1 —5 —I1 Sl_1 og | +5| o 9 | 07/26/2021 ELECTRICAL CIRCUITS The End ° ELECTRICAL CIRCUITS Nodal voltage analysis ° Nodal voltage analysis may be applied to both d.c. and a.c. circuits. Nodal analysis * A node of a network is defined as a point where two or more branches are joined. If three or more branches join at a node, then that node is called a principal node or junction. * Anode voltage is the voltage of a particular node with respect to a node called the reference node. ° If, for example, node 3 is chosen as the reference node then V,; is assumed to mean the voltage at node 1 with respect to node 3. 07/26/2021 ELECTRICAL CIRCUITS 12 The reference node is at point A ViaVe Vi Vin 8—¥s At node 1, = BV 1 6 5 50 — r__] CO ie., 1.367V; — V; —0.2V;—1.6=0 62 A 4 oo 3 V, Vo-V,; Vo-V At node 2, 2,2 t,o Sug 2 l 3 12 2Q ie, -V, +1.833V, — 0.333V; +0=0 2 32 V; V3;-V> V3+8-V; Atnode 3, —+— 4+—+—_ =0 4 3 5 ie, —0.2V, — 0.333V>+0.783V3+1.6=0 07/26/2021 ELECTRICAL CIRCUITS 15 Vy —V2 = —0.2 —16|] [1.367 —02 —16 1.833 —0.333 0 -1 -0333 0 —0.333 0.783 1.6 -0.2 0.783 16 V3 -1 ~ 11367 —1 —1.6) 11367. —!1 —0.2 -1 1.833 0 -1 1,833 —0,333 -0.2 0.333 1.6 0.2 —0.333 0.783 07/26/2021 ELECTRICAL CIRCUITS 16 Solving for V, -V> Solving for V> gives: — 1.6(—0.8496) + 1.6(—0.6552) —l ~ 1.367(1.3244) + 1(—0.8496) — 0.2(0.6996) -V -l 0.31104 hence > from which, voltage, V, = 0.31104 = 0.82093 0.82093 = 0.3789 V Thus the current in the 2 @ resistor = 2 = = = 0.19 A, flowing from node 2 to node A. 07/26/2021 ELECTRICAL CIRCUITS 17