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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.
