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Could someone please explain to me why is part b ii) the total resistance of the cable is 2 ohms or rather why the resistance if the cable is not given by multiplying 64 (the resistance of one wire) by 32 (the number of wires in the cable). Sorry about the formatting. enter image description here

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    $\begingroup$ The wires are in parallel, not in series. $\endgroup$ – Farcher Oct 31 '19 at 11:12
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The cable here consists of wires connected in parallel. This means that instead of current passing through 1 wire it can pass through all 32 of them, you can imagine this as if there were hypothetical wires before with infinite resistance but then you changed them with wires with finite resistance, this means the total resistance must decrease.

You can either use the rule for resistance of parallel connections $\frac {1} {R_T}=\frac {1} {R_1} + \frac {1} {R_2} + \frac {1} {R_3} + .....$

Or perhaps a more intuitive approach would be using conductivity: since there are 32 identical routes for the current to pass in (in parallel) so the conductivity is multiplied by 32 and thus the resistance is divided by 32.

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While the "resistors in parallel" explanation works,
it might be good to think about the geometrical factors in the resistance of a cylindrical conductor as $$R=\rho \frac{L}{A}$$

So, for one of the given wires, $$(64\ \Omega)=R_1=\rho_1 \frac{L_1}{A_1}.$$

The given arrangement of the 32 identical wires

  • keeps resistivity $\rho$ [=1/conductivity] unchanged (since it's the same material): so $\rho=\rho_1$
  • keeps length L unchanged: so $L=L_1$
  • but increases the effective cross-sectional area by a factor of 32: $A=32A_1$

So, $$R=\rho\frac{L}{A}=\rho_1\frac{L_1}{(32 A_1)}=\frac{1}{32}R_1=\frac{1}{32}(64\ \Omega)=2\ \Omega$$

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