In sensor signal conditioning, a Wheatstone bridge converts small resistance changes into a voltage. The bridge output is zero when the ratio R1/R2 equals the ratio R3/Rx.

• A. The bridge output is zero when R1/R2 equals R3/Rx.
• B. The bridge output is always nonzero unless all four resistors are equal.
• C. The bridge output is a current, not a voltage.
• D. The bridge output depends on the absolute values, not the ratios.

Answer: (A)

The essential idea is that a Wheatstone bridge balances when the two divider networks on opposite sides have the same ratio, so the mid-point voltages are equal and no differential output appears. In this setup, the left leg is formed by R1 on top and Rx on the bottom, while the right leg uses R2 on top and R3 on the bottom. The voltages at the midpoints are V_L = V_in · Rx/(R1 + Rx) and V_R = V_in · R3/(R2 + R3). The bridge output is the difference V_L − V_R. Setting this difference to zero gives Rx/(R1 + Rx) = R3/(R2 + R3). Cross-multiplying leads to R1·Rx = R2·R3, which rearranges to R1/R2 = R3/Rx. That is exactly the condition stated, so the bridge output is zero when those ratios are equal.

This is the practical basis for sensor signal conditioning: a small change in Rx (the sensor) unbalances the bridge slightly, producing a measurable voltage proportional to the change. The other options don’t fit because a balanced bridge can occur without all four resistors being equal, the output is a voltage difference (not a current), and the balance depends on the ratios rather than the absolute resistor values (aside from overall scale and tolerance effects).