How much electricity would be reduced in a 263 milimetres wire from 2.0 volts?
To answer this question fully, we would have to know something
about the wire, such as the amount of resistance per meter. Knowing
this we can calculate the ohms per millimeter. Next we need to know
what other loads are in the circuit, such as resistors. All taken
together we would have a series circuit. From Ohms Law, we know the
current is constant in a series circuit. Therefore the current
through the wire will be the same as the current flowing through
all loads. If the wire is connected directly across the voltage
source, the voltage drop will across the wire will be 2.0 volts,
the same as the source voltage. The current through the wire can be
calculated as follows: Assume wire resistance is 0.4 ohms per
kilometer. R = 263mm *(1Km/1,000,000mm) * (0.4 ohms / Km) =
0.000105 ohms I = V(source) / R(wire) I = 2.0 / .000105 I = 19,048
amps , if the source can supply it. A small gauge wire will get
hot, a fuse designed to protect the circuit will blow, or the
battery will quickly deplete itself.