this is a phy alvel qn:
1.the resistivity of the human body is low compared with the resistivity of skin, which is about 3.0x10^4 ohm m for dry skin. a person, who is well earthed,accidentally grabs a wire of diameter 0.40cm at a potential of 50v. his hand makes contact with the whole circumference of the wire over a distance of 9.0cm. the average thickness of the skin of his hand is 1.0mm.
estimate the current through the person(1.88mA). referring to the above scenario, discuss 2 factors which affect the magnitude of the current and hence affect the possible danger from electric shock.one obvious safety precaution is to keep live wires well insulated.what other safety precautions do you suggest?
2.how to draw R-V graphs for filament lamp and semiconductor diode?
3. for an electron, where does it have the highest potential?(there is one part of the circuit which is earthed as well as some resistors, but i don't know how to draw it out here)
4. a power station generates a fixed power at a voltage V. The power is transmitted along cables of total resistance R. show that the power loss as heat in the cables is proportional to (1/v^2). this in turn suggests that the power loss in the cables should be reduced to almost zero by a suitable choice of voltage V. discuss whether is the approach practical.
urm, anyone?
1) to find the current, you need to know the voltage and resistance for the path it travels. To find resistance, you know the area of contact ( diameter -> circumference, times the distance gives you an area.) The length at which current flows ( thickness), and resistivity, R = p (l/A) done. 50 v --> current obtained
if you look a the equaltions, putting R into I = v/r , length of current is unchanged, unless you slice your skin, resistivity can be considered, if the hand is wet, but its assumed constant here. the ones that will affect is voltage, area of contact, ie diameter of wire, and how much the guy stupidely wants to place his hand on the wire.
precaution, dont use wet hands, dont let chewy pets near insulated wires(like hamsters), dont touch what you dont know.
2)compare the V, I graphs from those (should be in text books) get the gradient at a few sensible points, compare with voltage at that point. Is it an increaing trend? make a rough estimated sketch. Note, the graph is likely to be non linear, as with most things in real life. Lots of factors, if the original VI graph tapers off something, probably the Reistance graph leads towards zero or infinity, in a non linearish way.
3)the highest electric potential should originate at the source, eg 12V+ battery end, and 0V is at earth. i cant remember if the electron reference makes it negative or what, but im sure the earthing point is defined as 0. I dunno how the diagram looks like, but the highest potential point should be the greatest 'distance' from the earth, behind the largest no of resistance that drains away the potential. For the electron the highest potential is the negative terminal. Its like the highest potential is at the top of a hill, and the resistors are are slopes, and the earth is flat ground. See if you can trace a path or something.
4)This is the standard proof for power loss thing that physics questions like to base on. Power = V I = I^2 R
power loss is alawys in terms of I^2 R, as it is dependent on resistance and the amt of charge flowing per time
Power loss = I^2 R , but since current is a function of the power generated at the station, where I = P(generated)/V (from P = VI)
we have Power loss = (P(generated)/V)^2 R hence, you can the power loss and V^2 relation.
Do not use Power loss = VI to explain it, as V is not known. The V here is voltage over the tranmission line, not V generated at the station.(much like voltage over a resistor, not the emf of a battery) You only know current through it, and its resistance from the given information. Current can be obtained from power and voltage at the station.
Hope this helps
Hi,
For Q1, the tricky thing I thought is to correctly identify length l and area A.
l refers to the thickness of skin (0.001m).
A refers to the surface area of the wire (2 pi x 0.002 x 0.09 sq m).
Thanks.
Cheers,
Wen Shih
For number 4, it is not practical not because of reasons due to equations.
It is not practical because you need to have step-up and step-down transformers to change the voltage. Being inefficient, transformers will have power losses as well. The question is whether suffering a power loss from these transformers to reduce the power loss in wires to near zero is worth it.
cheers ,that hardcore electrical engineering.. a module i hate when i study in diploma in electronics and comp engineering