How much current will flow through the $1\Omega$ resistor in the circuit
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Always check for the balanced Wheatstone bridge condition ($R_1/R_2 = R_3/R_4$) in complex-looking resistor networks. It often simplifies the problem by allowing you to remove the central branch.
Step 1: Identify the arrangement as a Wheatstone bridge circuit, with the \(1\ \Omega\) resistor connected between the two midpoints of the bridge.
Step 2: Check the balance condition for the bridge, when the ratio of resistances in one pair of arms equals the ratio in the other pair, the bridge is balanced.
Step 3: In a balanced bridge, both midpoints sit at exactly the same electric potential, so the potential difference across the connecting resistor is zero.
Step 4: Apply Ohm's law across the connecting resistor with zero potential difference: \( I = V/R = 0/1 \).
\[ I = \boxed{0\ \text{A}} \]
