Question 705065

Given unstrained values $R = R' = 100\Omega$.

For the strained condition, the resistors become $R + \Delta R$ and $R' - \Delta R$.

The bridge output voltage is $V_{ab} = V_a - V_b$.

With $V = 10$V, $V_a = \frac{10(R + \Delta R)}{(R + \Delta R) + (R' - \Delta R)} = \frac{10(R + \Delta R)}{R + R'}$ and $V_b = \frac{10(R' - \Delta R)}{R + R'}$.

Thus, $V_{ab} = \frac{10}{R+R'} (R + \Delta R - R' + \Delta R) = \frac{10}{200} (2\Delta R) = \frac{\Delta R}{10}$.

Given $V_{ab} = 0.3$ V, we have $\frac{\Delta R}{10} = 0.3$, which implies

$\Delta R = 3\Omega$.