First slide

combination of resistance

Question

If a rod has resistance 4 $Ω$ and if rod is turned as half cycle then the resistance along diameter

Difficult

A

1.56 $Ω$
B

2.44 $Ω$
C

4 $Ω$
D

2 $Ω$

Solution

Suppose the length of the rod = /. Given that its resistance is 4 $Ω$
$∴$ Resistance per unit length $= (4 / l) Ω$
The half circle of turned rod is shown in fig.
From figure $πr + r + r = l$
or $r = \frac{l}{(2 + π)}$
Its diameter $= 2 r = \frac{2 l}{(2 + π)}$
Now, resistance along diameter $= \frac{2 l}{(2 + π)} \times \frac{4}{l}$
$= \frac{8}{(2 + 3 \cdot 14)} = 1 \cdot 56 ohm$

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