A transformer is plugged into a regular U.S. Wall outlet. There are 200 turns in the...

A transformer is plugged into a regular U.S. Wall outlet. There are 200 turns in the primary coil, and 4000 turns in the secondary coil. A) What is the rms voltage and peak voltage on the secondary coil? (two answers) B) An electrical worker accidently grabs one of the loops of the secondary coil with a single bare hand. Her other hand is in midair, and both of his shoes are firmly on the ground. The resistance in her body is negligible compared to his shoes. Each shoe has a 2mm-thick rubber sole, with a sole-area of 0.02 m2 . The rubber has a resistivity of 1 MW•m. What is the rms current that runs through the worker's torso? (Warning: there are two shoes.) Hint: A drawing is almost certainly helpful. C) There is a 50-Amp fuse on the wall outlet. Will this fuse trip before the full current (calculated in part B) passes through the worker's body? (Careful: There is a complication here, since the fuse is on the primary-coil side of the transformer.).

Solutions

Expert Solution

a) voltage input = regular U.S= Vprimary =120V /this is rms

numer turns in the primary coil Nprimary =200

numer turns in the secondary coil Nsecondary =4000

Vsecondary

apply trnsformer equation Vprimary /Vsecondary =Nprimary /Nsecondary

120 / Vsecondary =200 / 4000

rms Vsecondary =120x40 /2 =2400 volt

peak voltage of the Vsecondary = Vsecondary = x2400 volt =3394.11 volts

b) resistance of his shoe can be find by equation R = resistivity = 1 mega ohm =106 ohm

length L =2mm=2x10-3 m area A = 0.02m2

R = 106 x2x10-3 /0.02 =100000 ohm for one she

boath shoe are in parallel so effective resistance to voltage = 100000 x100000 / (100000 +100000 ) =50000ohm

so current flow through his body = working voltage / effective resistance =2400 volt /50000ohm =0.048ampere

c) 50 A fuse in primary

the current flow in primary while worker ketting shock is

apply transformer equation Vprimary /Vsecondary = Isecond / Iprimary -----/ Isecond =0.048ampere Vprimary =120

Vsecondary= 2400 volt

Iprimary = 0.048 x2400 / 120 =0.96 amps

fuse will not burn out because current flowing is lessthan the fusing current (0.96A < 5A)

genius_generous answered 1 year ago

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