The resistance of the conductor in a strain gauge increases when a force is applied to it - Leaving Cert Physics - Question Question 1 - 2014

To find the resistance of the unknown resistor, we can apply the principle of the Wheatstone bridge, which states that:

$R_1 / R_2 = R_3 / R_4$

Substituting the known values:

$5.1 \, ext{Ω} / 11.9 \, ext{Ω} = R / 40.5 \, ext{Ω}$

Cross-multiplying gives:

$R = (5.1 \, ext{Ω} / 11.9 \, ext{Ω}) \times 40.5 \, ext{Ω}$

Calculating this, we find:

$R \approx 17.36 \, ext{Ω}$

The resistance (R) of a wire can be expressed by the formula:

$R = \frac{\rho L}{A}$

Where:

• (R) is the resistance,
• (\rho) is the resistivity of the material,
• (L) is the length of the wire,
• (A) is the cross-sectional area of the wire.

The area (A) can be expressed in terms of diameter (d):

$A = \frac{\pi d^2}{4}$

When the radius of the wire is doubled, the cross-sectional area increases by a factor of four, because:

$A \propto r^2$

Thus, the new area (A') becomes:

$A' = \pi (2r)^2 = 4\pi r^2 \\ = 4A$

Since resistance is inversely proportional to cross-sectional area, the new resistance (R') will be:

R' = \frac{\rho L}{A' = \frac{\rho L}{4A} = \frac{R}{4}

Thus, the resistance decreases by a factor of 4.