Conducting loop

Assume the wire's resistance is 2 ohms. Then the voltage of AB is 3 volts by ohm's law which is V = IR. When the same wire is bent into a circle, the distance of AB is half of the length it used to be therefore the resistance also halfs. Now each branch is 1 ohm and the voltage drop across each branch is still 3 volts by using the voltage divider formula. Subsistuting these values in the ohms law equation, we solve for I = current. 3= Current * 1. Now its simple to deduce that current flowing across point C is 3 amps.

A battery is connected to the Point A and Point B of a homogenous wire. Therefore, both flow together and it is bent into a circle shape so both of the wires are at the same length. Then find Point C I = 1.5A Amps of Point C = Point A + Point B Amps of Point C = 1.5 + 1.5 = 3 Amps

This one is extremely simple.

Alright so we know $I = \frac{V}{R}$ (ohm's law), and the internal resistance of the wire can be given as R = * $\frac{p \times L}{A}$ , where * V is the voltage, I is the current, R is the resistance, L is the length of the conducting wire, p is the resistivity of the conducting wire, and A is the perpendicular-cross sectional area of the wire (imagine the wire being a long cylinder, and A being the area of the base circle).

From these, we can plug in the resistance formula and achieve this...

*I ∝ * $\frac{1}{L}$

which is a bit counter-intuitive, but hey, you've probably already been warned about this.

This means that if the length of wire from one battery to another is shortened by a factor of $\frac{1}{2}$ , the current running through will raise by a factor of $2$ .

This gives us our answer, $2 \times 1.5 Amps = \boxed{3.0 Amps}$

Let the resistance of the wire be $R$ . The first thing to remember is that the resistance is directly proportional to the length of the wire . So, in the second case, as the semicircular piece of the wire between $A$ and $B$ is exactly half the length of the original wire, the resistance of each piece will be $\frac{R}{2}$ . As the two pieces of the wires are connected in parallel, the voltage across the terminals will be the same, say $V$ . So, by Ohm's Law, $V = I'\frac{R}{2} = IR$ . Since the resistance is halved, the current is doubled so that $V$ remains the same. Therefore, $I' = 2I = \boxed{3A}$

Its current in parallel.

current flows equally from both sides.

and let the current flowing from c is x

1/AB = 1/x + 1/x

1/1.5 = 2/x

x = 2 x 1.5

x=3