What will be the electrical output with this force if we have a four active gauges, each with a nominal 150 Ω resistance and a gauge factor of 2 and a 10 V power supply?
The correct answer and explanation is:
The electrical output from a strain gauge circuit can be determined using the concept of a Wheatstone bridge, where the strain gauges are typically arranged in a bridge configuration. The formula to calculate the output voltage (V_out) from a Wheatstone bridge is:
Vout=Vin×ΔRRV_{out} = V_{in} \times \frac{\Delta R}{R}
Where:
- VinV_{in} is the input voltage (10 V in this case),
- ΔR\Delta R is the change in resistance of the strain gauges due to the applied force,
- RR is the nominal resistance of the strain gauges (150 Ω).
Step 1: Strain Gauge Output Voltage
The strain gauges in the bridge configuration will change their resistance when subjected to strain. The change in resistance ΔR\Delta R is proportional to the applied strain (ε\varepsilon), and can be calculated using:
ΔR=R0×ε×GF\Delta R = R_0 \times \varepsilon \times GF
Where:
- R0R_0 is the nominal resistance (150 Ω),
- ε\varepsilon is the strain (this is not given explicitly, but it is implied the strain will cause the gauge to change resistance),
- GFGF is the gauge factor (2).
Since the strain is not given explicitly, an assumption is made that the output voltage will depend on the applied strain as it affects resistance.
Step 2: Wheatstone Bridge Calculation
In a full Wheatstone bridge with four active gauges (two in tension and two in compression), the output voltage will be maximized when the resistance changes in opposite directions for the two gauges. The output voltage is derived from the difference in resistance between the gauges.
Assuming the total change in resistance (due to strain) is small and the nominal voltage is applied across the bridge, the output voltage can be calculated as:
Vout=Vin4×GF×εV_{out} = \frac{V_{in}}{4} \times GF \times \varepsilon
Step 3: Estimation
Given the 10 V supply and a gauge factor of 2, the output voltage depends directly on the strain applied and can be calculated if the strain (ε\varepsilon) is known. If the strain is small, the voltage will also be small, and the output can be amplified using an op-amp circuit to achieve a usable signal for measurement.
This setup allows the measurement of small forces with high accuracy, depending on the sensitivity of the strain gauges and the bridge configuration.