Question 705065

SCQMEDIUM

In the circuit shown in the figure, the resistances $R$ and $R'$ change due to strain. While $R$ increases, $R'$ decreases by the same amount $\Delta R$ due to the applied strain. The unstrained values of $R$ and $R'$ are $100\Omega$ each. If same strain is applied to all the resistors, and the output voltage $(V_{ab})$ changes to $0.3$ V, then $\Delta R$ is closest to

(A)

$3\Omega$

(B)

$1.5\Omega$

(C)

$4.5\Omega$

(D)

$6\Omega$

Detailed Solution

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

$\Delta R = 3\Omega$ .

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