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Two mesh loops share a common middle branch. The left loop contains a…
Question
Two mesh loops share a common middle branch. The left loop contains a 6 V cell and a 1 ohm resistor; the right loop contains a 4 V cell and a 2 ohm resistor; the shared middle branch is a 2 ohm resistor. The two cells drive their respective loop currents (i1 in the left mesh, i2 in the right mesh) such that the middle branch carries (i1 - i2). Mesh equations: 6 = 1·i1 + 2(i1 - i2) and 4 = 2·i2 + 2(i2 - i1). Find the current through the middle (2 ohm) branch, in ampere.
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Step-by-step solution
Mesh 1: 6 = i1 + 2(i1 - i2) = 3 i1 - 2 i2.
Mesh 2: 4 = 2 i2 + 2(i2 - i1) = -2 i1 + 4 i2.
From mesh 1: 3 i1 - 2 i2 = 6.
From mesh 2: -2 i1 + 4 i2 = 4, i.e.
-i1 + 2 i2 = 2.
Adding the first to (this multiplied appropriately): from -i1 + 2 i2 = 2, i1 = 2 i2 - 2.
Substitute: 3(2 i2 - 2) - 2 i2 = 6 -> 6 i2 - 6 - 2 i2 = 6 -> 4 i2 = 12 -> i2 = 3, then i1 = 4.
Middle branch current = i1 - i2 = 4 - 3 = 1 A.
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