# 13 for mesh 2 applying kvl gives r2 i2 v2 r3 i2 i1

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Unformatted text preview: meshes 1 and 2. Although a mesh current may be assigned to each mesh in an arbitrary direction, it is conventional to assume that each mesh current ﬂows clockwise. As the second step, we apply KVL to each mesh. Applying KVL to mesh 1, we obtain The direction of the mesh current is arbitrary— (clockwise or counterclockwise)—and does not affect the validity of the solution. −V1 + R1 i1 + R3 (i1 − i2 ) = 0 or (R1 + R3 )i1 − R3 i2 = V1 (3.13) For mesh 2, applying KVL gives R2 i2 + V2 + R3 (i2 − i1 ) = 0 or −R3 i1 + (R2 + R3 )i2 = −V2 (3.14) Note in Eq. (3.13) that the coefﬁcient of i1 is the sum of the resistances in the ﬁrst mesh, while the coefﬁcient of i2 is the negative of the resistance common to meshes 1 and 2. Now observe that the same is true in Eq. (3.14). This can serve as a shortcut way of writing the mesh equations. We will exploit this idea in Section 3.6. The third step is to solve for the mesh currents. Putting Eqs. (3.13) and (3.14) in matrix form yields R1 + R3 −R3 −R3 R2 + R 3 i1 V1 = i2 −V2 The shortcut way will not apply...
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## This note was uploaded on 07/16/2012 for the course KA KA 2000 taught by Professor Bkav during the Spring '12 term at Cambridge.

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