Answer:
170/179 A ≈ 949.7 mA
Explanation:
Finding equivalent impedance of a bridge network can be simplified by converting one end of it from a Δ to a Y configuration. The equivalent resistance in the Y connection is the product of the two resistors connected to that node, divided by the sum of all three resistors.
Considering the left side of the top two branches, the equivalent Y network values are ...
(10)(15)/(10+15+25) = 150/50 = 3 . . . ohms
connected to the right-end node
(25)(10)/50 = 5 . . . ohms
connected to the top-center node, and ...
(25)(15)/50 = 7.5 . . . ohms
connected to the center node of the middle branch.
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Then the top two branches of this circuit have an equivalent impedance equal to 3 ohms in series with the parallel combination of (5+20) ohms and (7.5+10) ohms. That value is ...
3 + (25)(17.5)/(25 +17.5) = 226/27 ≈ 13.294 . . . ohms
This, in turn, is in series with the 2.5 ohms in the bottom branch, bringing the total circuit resistance to about 15.794 ohms, or 537/34 ohms.
The total source current is then ...
I = V/R = (15 v)/(537/34 Ω) = 170/179 A ≈ 0.9497 A
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Alternate solutions
Alternatively, you can write and solve three node or loop equations for this circuit. For example, the equations for CCW loop currents, with I1 as the bottom loop, could be written in augmented matrix form as ...
[tex]\left[\begin{array}{ccc|c}27.5&-10&-15&-15\\-10&55&-25&0\\-15&-25&50&0\end{array}\right][/tex]
The result is the same.
