Question

A strain gauge has a nominal resistance of $600\ \Omega$ and a gauge factor of 2.5. The strain gauge is connected in a DC bridge with three other resistances of $600\ \Omega$ each. The bridge is excited by a $4\text{ V}$ battery. If the strain gauge is subjected to a strain of $1\ \mu\text{m/m}$, the magnitude of the bridge output will be}

• A. zero
• B. $2.50\ \mu\text{V}$
• C. $500\ \mu\text{V}$
• D. $750\ \mu\text{V}$

Hint:

For a quarter bridge with equal resistances, the nominal value of the resistance (here $600\ \Omega$) does not affect the output voltage.
Only the excitation voltage, gauge factor, and strain determine the output.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem requires finding the output voltage of a Wheatstone bridge with one active strain gauge (quarter-bridge configuration).

Step 2: Key Formula or Approach:
For a quarter-bridge configuration where one arm contains an active strain gauge of resistance $R + \Delta R$ and the other three arms contain equal fixed resistances $R$:
\[ V_{out} = \frac{V_{in}}{4} \left( \frac{\Delta R}{R} \right) \]
The fractional change in resistance is related to strain $\varepsilon$ by the gauge factor $GF$:
\[ \frac{\Delta R}{R} = GF \cdot \varepsilon \]
Thus, the bridge output is:
\[ V_{out} = \frac{V_{in}}{4} \cdot GF \cdot \varepsilon \]

Step 3: Detailed Explanation:

• Given parameters:
Nominal resistance, $R = 600\ \Omega$.
Gauge factor, $GF = 2.5$.
Excitation voltage, $V_{in} = 4\text{ V}$.
Applied strain, $\varepsilon = 1\ \mu\text{m/m} = 1 \times 10^{-6}$.

• Substitute these values into the bridge output voltage formula:
\[ V_{out} = \frac{4\text{ V}}{4} \times 2.5 \times (1 \times 10^{-6}) \]
\[ V_{out} = 1 \times 2.5 \times 10^{-6}\text{ V} = 2.5 \times 10^{-6}\text{ V} \]

• Converting to microvolts ($\mu\text{V}$):
\[ V_{out} = 2.50\ \mu\text{V} \]

Step 4: Final Answer:
The magnitude of the bridge output voltage is $2.50\ \mu\text{V}$, which corresponds to Option (B).