In a balanced Wheatstone bridge, the current through the central resistor is zero. Therefore, the \( 5 \, \Omega \) resistor can be disregarded. The circuit can be simplified by its removal.
Step 1: Circuit Simplification The \( 5 \, \Omega \) resistor is removed. The remaining resistors are combined. The two \( 3 \, \Omega \) resistors in the top branch are in series, yielding a combined resistance of: \[ R_{\text{top}} = 3 + 3 = 6 \, \Omega. \] The two \( 6 \, \Omega \) resistors in the bottom branch are in parallel, resulting in an equivalent resistance of: \[ R_{\text{bottom}} = \frac{6 \cdot 6}{6 + 6} = \frac{36}{12} = 3 \, \Omega. \] These two resistances, \( R_{\text{top}} \) and \( R_{\text{bottom}} \), are now in parallel, giving: \[ R_{\text{eq}} = \frac{6 \cdot 3}{6 + 3} = \frac{18}{9} = 2 \, \Omega. \] Finally, the \( 2 \, \Omega \) resistor is added in series with \( R_{\text{eq}} \): \[ R_{\text{total}} = R_{\text{eq}} + 2 = 2 + 2 = 6 \, \Omega. \]
Step 2: Ohm's Law Application The total current is calculated using Ohm's Law: \[ I = \frac{V}{R_{\text{total}}}. \] Substituting \( V = 6 \, \text{V} \) and \( R_{\text{total}} = 6 \, \Omega \): \[ I = \frac{6}{6} = 1 \, \text{A}. \]