Find change in resistance for mercury-tube breathing monitor

[JJaX]

AP Physics Help--Emergency

My physics teacher sometimes gives out rediculously difficult worksheets. I'm having trouble with one problem in particular:

A breathing monitor girds a patient with a mercury-filled rubber tube and measures the variation on tube resistance. The tube has an unstretched length of 1.25 m and inside diameter of 2.51 mm. The monitor is connected to a 100-mV power supply, and the total resistance of the circuit is that due to the mercury plus 1.00 Ω (an internal resistance of the power supply). Determine the change of current through the monitor as the patient draws a breath and stretches the hose by 10.0 cm. Take ρ(Hg) = 9.40 x 10^(-7) Ω ◦ m.

Any help would be greatly appreciated.

 

[Aaron Barnard]

The change in the current through the monitor is directly proportional to the change in resistance.

The change in resistance is caused by the change of length in the tube due to the patient's breath. The formula for calculating this change is ΔR = ρL (ΔL/L), where ρ is the resistivity of the material, L is the original length, and ΔL is the change in length.

Plugging in the given values, we have:

ΔR = (9.40 x 10^(-7) Ω ◦ m)(1.25 m)(10.0 cm/1.25 m) = 7.52 x 10^(-6) Ω

Now, the total resistance of the circuit is the sum of the internal resistance and the change in resistance, so

R = 1.00 Ω + 7.52 x 10^(-6) Ω = 1.0075 Ω

Now, using Ohm's law, we can calculate the change in current:

I = (100 mV)/(1.0075 Ω) = 99.92 mA

Therefore, the change in current through the monitor is 99.92 mA.

 

[M98Ranger]

To find the change in resistance for the mercury-tube breathing monitor, we can use the formula for resistance, R = ρL/A, where ρ is the resistivity of the material, L is the length of the tube, and A is the cross-sectional area of the tube.

First, we need to find the initial resistance of the mercury-filled rubber tube. We can calculate the cross-sectional area using the formula A = πr^2, where r is the radius of the tube. The radius can be found by dividing the inside diameter by 2, so r = 1.255 mm = 0.001255 m.

Using this value for the radius, we can calculate the initial cross-sectional area as A = π(0.001255 m)^2 = 4.94 x 10^-6 m^2.

Next, we can calculate the initial resistance of the tube using the formula R = ρL/A. Plugging in the given values, we get R = (9.40 x 10^-7 Ω ◦ m)(1.25 m)/(4.94 x 10^-6 m^2) = 0.238 Ω.

Now, we need to find the change in resistance when the tube is stretched by 10.0 cm. We can use the same formula, but with a new length of 1.35 m. This gives us a new resistance of R = (9.40 x 10^-7 Ω ◦ m)(1.35 m)/(4.94 x 10^-6 m^2) = 0.256 Ω.

To find the change in resistance, we can subtract the initial resistance from the final resistance. This gives us a change in resistance of 0.256 Ω - 0.238 Ω = 0.018 Ω.

Finally, we can use Ohm's Law (V = IR) to find the change in current through the monitor. Since the power supply is connected to a 100-mV power supply, we can use this voltage to find the change in current. We can rearrange the formula to solve for current, I = V/R. Plugging in the values, we get I = (0.100 V)/(0.018 Ω) = 5.56 A.

So, the change in current through the monitor as the patient draws a breath