[GreenTeaBanana]

sorry i dont know why its huge
I know its bad for me to ask for help here but im pretty desperate...
Im more interested in a good explanation because i honestly dont get physics and im really scared im going to fail my final...
what you are looking at is a test that i recently got a 52 on...and im not proud of it
I really need someone to help me understand this pleaseee ):

[Voltage]

from the initial conditions, we see that the equivalent resistance of the wire Rwire = Voltage(battery) / I(measured) = 10 ohms

Question 28: We have effectively 2 wires or 2 resistances in parallel
1/Requivalent = 1/R1 + 1/R2
R1R2/(R1+R2) = 5
so the equivalent resistance seen by the battery is 5 ohms giving a current of Vbattery/Requivalent = 2 Amps

Question 29: The equivalent resistance is now two wires in series.
Requivalent = R1 + R2 = 20 ohms
I = V/Requivalent = 1/2 Amp

Question 30: We can divide the equivalent resistance of one wire into 2 parts (2 resistances in series) so that each half is 5 ohms
So now we have a resistance (5 ohms contributed by half of R1) added to the equivalent resistance of half R1 in parallel with R2
Requivalent = 1/2*R1 + (1/2)R1*R2/[(1/2)R1 + R2]
= 5 + 5*10/(5+10) = 25/3 Ohms

I = V/Requivalent
=30/25 = 6/5 Amps or 1.2 Amps

[GreenTeaBanana]

ahhh thank you soo muchh

Can you also help me with this problem? ): ahhh ignore my scribbles
if i say that 38 is c because an electric field goes from positive to negative and the vector for c is a straight line...would my thought process be correct? or am i way off?
i dont get 39 and 40 though :/

[Voltage]

I think you should start talking to whoever is teaching you because these questions are testing foundational material at a very basic level.

Anyways, I'll refrain from giving away the answers since that won't really help you on your exams. Think about electric fields in general as they apply to point charges. The summation of each field on a point charge determines the electric field at the given point. To clarify, think about the -3 micro-Coulomb charge and how it creates an electric field around itself. Then think about the 15 micro Coulomb charge. The point X is being acted on by both charges. By convention, electric field strengths and directions have positive sign which means that the magnitude of the electric field tells you how strong it is based on what a theoretical positive point charge would do in this field. So pretend X is a positive point charge. It has one vector pushing it to the left since positive-positive charges repel. On the other side, the charge has an attractive vector that points to the left as well proportional to the distance squared. So we can think of one really long vector to the left(contributed by the 15microC) and another smaller vector pointing to the left(contributed by the -3 microC). SO your thought process does work for that question, but may not work for other questions. Thinking about the charges acting independently then summing them all together is the way to go. Anyways the answers your marked on there appear to be correct.