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AS Physics P1 MCQs Preparation Thread.

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please help!
first we find the resistance of the wires
0.005 x 800 =4 and because there are 2 wires which forms the loop so 4x2 = 8
now we find the V dissipated by the resistance so V= IR = 8x0.6 = 4.8V
this means that 4.8 will be lost during transmission
and the relay requires 16, so 16+4.8 = 20.8
 
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first we find the resistance of the wires
0.005 x 800 =4 and because there are 2 wires which forms the loop so 4x2 = 8
now we find the V dissipated by the resistance so V= IR = 8x0.6 = 4.8V
this means that 4.8 will be lost during transmission
and the relay requires 16, so 16+4.8 = 20.8
Excuse me 6Astarstudent... did u take ur physx p1 just now?
If so, and if u have the temperament to do so, can u kindly PM me the answers u remmbr?
 
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first we find the resistance of the wires
0.005 x 800 =4 and because there are 2 wires which forms the loop so 4x2 = 8
now we find the V dissipated by the resistance so V= IR = 8x0.6 = 4.8V
this means that 4.8 will be lost during transmission
and the relay requires 16, so 16+4.8 = 20.8

thanks buddy, i had forgotten about the other wire
 
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