Consider a copper wire 1 mm in diameter providing the power needed to run an appliance drawing a 4.8 kW at 12 V. Assuming that no heat is radiated away from the wire while the current flows:A. What will the temperature of the wire be after the current has run for 1 second through the wire?B. What will the physical condition of the wire be at that time?(The wire was initially at 20 degrees C)
Battery's have their very own resistance (noted as their 'inner resistance'). the cost of the indoors resistance relies upon on the form of battery and its age. the full resistance of your circuit hence would be plenty greater effective than 0.064 ohm, so the present would be decrease than ninety 3.75A.. you will no longer get a electric marvel from 6V (although your tongue can detect it - no longer recommneded!). approximately 30mA around the guts can quit it. yet you may choose a plenty bigger voltage than 6V by way of resistance of the physique. in spite of the shown fact that there is achieveable of overheating the battery and/or the twine, which includes achieveable of hearth in some circumstances. Please do no longer attempt this with AC mains. There are severe detrimental aspects of deadly electric marvel and beginning a hearth. The kin risk-free practices equipment (fuse/circuit breaker/RCD) are meant to shrink the skill if the present is basically too severe or if there is an earth-leak - yet you are able to no longer possibility the risk-free practices of your self and others in this kind.
Assume the wire is 1 meter long. Resistance of a wire in Ω R = ρL/A ρ is resistivity of the material in Ω-m L is length in meters A is cross-sectional area in m? A = πr?, r is radius of wire in m resistivity Cu 17.2e-9 Ω-m or 17.2e-6 ohm-mm R = (17.2e-6)(1000) / π(0.5)? = 0.022 ohms Appliance current is 4800/12 = 400 amps Power is I?R = 400? x .022 = 3520 watts. Problem, this power is at a level almost as high as that of the appliance. In other words, the wire is too thin to sustain that power level. .
