Physics A: Problem Set 7: Electric Resistance
|Barron's Let's Review:||9.4–9.5 Resistance|
|Wikipedia:||Electrical resistance and conductance, Ohm's law|
|HyperPhysics:||Resistance, Electric Shock|
|Khan Academy:||Circuits (Part 1)|
- An electric current of 1 mA through the human body is just barely detectable. At 5 mA, the sensation becomes painful. A current above 15 mA is sufficient to cause paralysis in major voluntary muscles. At 75 mA, the current will begin to scramble the normally coordinated contractions of the heart — a condition that is fatal if it persists for more than a few seconds. This information is summarized in the table below along with some information on the arm-to-arm resistance of the human body.
Physiological effects of current current (mA) resistance (Ω) sensation 01 dry skin 100,000 pain 05 wet skin 005,000 paralysis 15 fibrillation 75
- What voltage across the arms of a human with dry skin would result in a barely detectable current?
- The voltage you calculated in part a. is now applied to a human with wet skin. What is the effect?
- Why is it dangerous to handle electrical equipment with wet skin?
- What minimum voltage is needed to produce sensation in wet skin?
- How could the situation described in part d. be demonstrated? (No exotic or dangerous equipment, please.)
This is a question about Ohm's law.
Use the appropriate values from the table. Be aware that the currents are stated in milliamps, not amps.
V = IR
V = (0.001 A)(100,000 Ω)
V = 100 V
Use the voltage calculated in part a with the resistance of wet skin in the table.
I = V = 100 V R 5000 Ω I = 0.020 A = 20 mA
A 20 mA current is more than enough to cause paralysis, but not quite enough to kill.
Since wet skin has a lower resistance than dry, the current will be greater for a given voltage and the physiological effect on a person will be worse.
Pick the appropriate numbers from the table, substitute, and solve.
V = IR
V = (0.001 A)(5000 Ω)
V = 5 V
The obvious answer is to find a 5 V battery and then touch it to wet skin. But 5 V batteries don't really exist (or at least, I've never seen one). We could take four 1.5 V batteries line them up end to end and create a 6 V battery, but there's a better answer — one that's more fun. Find a 9 V battery and lick the terminals. (I'm talking about the rectangular batteries with the two terminals on the top — the kind of battery occasionally found in smoke alarms, radio-controlled cars, and some alarm clocks.) The terminals are nice and close, which makes this an easy thing to do. Since 9 V is greater than the 5 V computed in part d, and since the distance between the terminals is much shorter than the distance between arms, and since the tongue is wet, the effect is more than just minimal sensation. It's distinctly uncomfortable — uncomfortable, but not dangerous. (Do not modify this experiment in any way. Do not eat the 9 V battery. Do not replace the 9 V battery with a car battery or an outlet. Do not inhale the battery. Do not put on a Superman costume and attempt to fly.)
- Which device in your home has the highest resistance: a flood light, a reading lamp, or a night light? All three devices are plugged into standard North American 120 V outlets.
Ohm's law says that resistance and current are inversely proportional. The highest resistance goes with the least current — the tiny little night light.
- You are a cruel and unusual electrical engineer and have been given the cruel and unusual task of designing an electric chair for the purposes of executing criminals in a cruel and unusual fashion.
- How is your victim to be attached to the chair and what special preparations would you do to ensure the chair worked most effectively?
- What minimum voltage is sufficient to achieve your task?
- What do the terms DC and AC refer to and which of these was used in electric chairs? Explain the reasoning behind this choice.
Attach the condemned to the chair with large metal electrodes to ensure a large current. Attach these electrodes to distant parts of the seated body — the top of the head and an ankle. Pad the metal electrodes with cloth or sponge to ensure a good connection between the hard metal of the electrodes and the soft flesh of the human body. Soak the padding in saltwater or some other conducting liquid to reduce its resistance. Shave the condemned's scalp to remove the insulating layer of hair.
Use Ohm's law with enough current to stop the heart (75 mA) and a whole body resistance for wet skin (5,000 Ω).
V = IR
V = (0.075 A)(5,000 Ω)
V = 375 V
DC is direct current — current that only flows in one direction. AC is alternating current — current that reverses direction periodically. AC is more lethal than DC at the same voltage, which is why AC was used in electric chairs.
- Given a wire with a resistance R, what will be the new resistance if…
- the wire is cut in half and only one half is used to conduct electricity,
- the wire is folded in half and both halves are used to conduct electricity?
Use the equation for the resistance of a wire.
R = ρℓ A
Cutting the wire in half reduces its length by a half and changes nothing else. Resistance and length are directly prorportional, so the new resistance is ½R.
(½R) = (½ρ)(ℓ) (A)
Folding the wire in half cuts the length in half, but doubles the area. (There are now two parallel conducting paths.) Resistance and length are still directly prorportional (which halves the resistance), but resistance and area are inversely proportional (which also halves the resistance). A half of a half is a quarter. The new resistance is ¼R.
(¼R) = (½ρ)(ℓ) (2A)
- A tungsten rod and an aluminum rod have the same length and resistance.
- What is the ratio of the cross sectional area of the tungsten rod to the aluminum rod?
- What is the ratio of the diameter of the tungsten rod to the aluminum rod?
Start with the equation for the resistance of a wire.
R = ρℓ A
Set resistance equal to resistance. (Al is the symbol for aluminum and W is the symbol for tungsten.)
RAl = ρAlℓ = ρWℓ = RW AAl AAl
Cancel the length from both sides, since it's the same for both wires.
ρAl = ρW AAl AW
Rearrange it into what we want — a ratio of areas.
AW = ρW AAl ρAl
The ratio of the areas is the same as the ratio of the resistances. There are other ways to derive this relationship using mathematical reasoning. You just saw how I did it today. Next up: numbers in, answer out.
AW = 52.8 × 10−9 Ωm AAl 26.5 × 10−9 Ωm
The ratio of the areas is basically 2:1.
The equation for the area of circle is famous.
A = πr2
It says that area is proportional to radius squared.
A ∝ r2
But wait, I hear some of you say, The problem didn't say radius, it said diameter. Don't we need to multiply or divide or do something with two somewhere? Well yes, if we were talking about an equation. Here all we care about are porportions.
If diameter is twice radius…
D = 2r
then diameter is directly proportional to radius.
D ∝ r
When two quantities are directly porportional, whatever change happens to one happens to the other. Thus…
A ∝ D2
D ∝ √A
Multiplying an area by 2 is the same as multiplying a diameter by √2. So, the ratio of the diameters is √2:1.