Question

A Wheatstone bridge strain gauge transducer is constructed on a diaphragm in such a way that when a force is applied on the diaphragm, the resistors \(R_1\) and \(R_4\) will be in compression, and the resistors \(R_2\) and \(R_3\) will be in tension. The bridge excitation voltage \((E_{\text{in}})\) is \(10\ \text{V}\). If all the resistors have a resistance of \(200\ \Omega\) in the absence of any force, and each resistance changes by \(20\ \Omega\) upon application of a force, what is the output voltage \(V_{\text{out}}\) (in Volts) from the Wheatstone bridge? (Round off your answer to the nearest integer.)

Hint:

- For a Wheatstone bridge with excitation top-to-bottom, \(V_{\text{out}}=E_{\text{in}}\!\left(\frac{R_2}{R_1+R_2}-\frac{R_4}{R_3+R_4}\right)\). - Opposite changes in adjacent arms (one increases, the other decreases) maximize bridge sensitivity.

Answer 1

Correct Answer: -1

Step 1: Assign the new resistances under load. Compression \(\Rightarrow\) resistance decreases, tension \(\Rightarrow\) increases. \[ R_1=200-20=180\ \Omega, R_4=200-20=180\ \Omega,
R_2=200+20=220\ \Omega, R_3=200+20=220\ \Omega. \] Step 2: Use the Wheatstone bridge divider relations (excitation across top and bottom nodes). Left mid-node potential: \[ V_L = E_{\text{in}}\frac{R_2}{R_1+R_2} = 10\cdot \frac{220}{180+220} = 10\cdot \frac{220}{400}=5.5\ \text{V}. \] Right mid-node potential: \[ V_R = E_{\text{in}}\frac{R_4}{R_3+R_4} = 10\cdot \frac{180}{220+180}=10\cdot \frac{180}{400}=4.5\ \text{V}. \] Step 3: Output voltage between the two mid nodes. \[ V_{\text{out}} = V_L - V_R = 5.5 - 4.5 = 1.0\ \text{V}. \] Rounded to the nearest integer: \(\boxed{1}\).