Question

Scenario: Voltage source is 15V. The Bridge is balanced and all devices are each at 400 Ohms. V is at 0V. What would the voltage output be if a Force increase to 200N causes the strain gauge sensor resistance to increase to 409 Ohms?

Answer 1

To solve the problem, we first need to understand that the bridge configuration mentioned (likely a Wheatstone bridge) produces an output voltage that is sensitive to changes in resistance.

When the bridge is balanced, the output voltage \( V_{out} \) is zero. The balance of a Wheatstone bridge can be expressed using the resistances in the four arms of the bridge:

- Let the resistances be:
- \( R_1 = R_2 = R_3 = R_4 = 400 \, \Omega \) (initially)

As the force increases, one of the resistances will change. In this case, the resistance of the strain gauge increases to:

- \( R_4 = 409 \, \Omega \) (the strain gauge)

### Bridge Output Calculation

In a Wheatstone bridge, the output voltage \( V_{out} \) can be calculated using the formula:

\[
V_{out} = V_s \cdot \frac{(R_2 / (R_2 + R_1)) - (R_4 / (R_3 + R_4))}{1}
\]

Where:
- \( V_s \) is the supply voltage (15V in this case)
- \( R_1 = 400 \, \Omega \)
- \( R_2 = 400 \, \Omega \)
- \( R_3 = 400 \, \Omega \)
- \( R_4 = 409 \, \Omega \)

### Plugging in the values

Using the input values:

- For the left side of the bridge:
\[
\frac{R_2}{R_1 + R_2} = \frac{400}{400 + 400} = \frac{400}{800} = 0.5
\]

- For the right side of the bridge:
\[
\frac{R_4}{R_3 + R_4} = \frac{409}{400 + 409} = \frac{409}{809} \approx 0.506
\]

### Now calculate \( V_{out} \)

Now substituting these values back into the output voltage formula:

\[
V_{out} = 15V \left( 0.5 - 0.506 \right) = 15V \cdot (-0.006) \approx -0.09V
\]

Thus, the voltage output \( V_{out} \) when the force exerts 200N and changes the resistance of the strain gauge to 409 Ohms is approximately **-0.09V**.

This indicates that the output is negative, which means the voltage would be measured below the reference point (0V).