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