Question

A 5 $\Omega$ constantan wire is bent to form a ring. Find the resistance across the diameter of the wire.

• A. $2.5\Omega$
• B. $1.25\Omega$
• C. $5\Omega$
• D. $0.625\Omega$

Answer 1

Answer: (B)

$1.25\Omega$

In parallel combination reciprocal of net resistance is equal to reciprocals of individual resistances present in parallel combination.

i.e $\frac{1}{R_{net}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}.......$

Across the diameter if we see there is two semicircles one is above and another is lower the diameter. The upper semicircle and lower semi circle have resistance 2.5 $\Omega$ each since length of semicircles are same and total resistance of the wire is 5$\Omega$

Both resistances are in parallel.
Hence,$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$R_{eq}=\frac{R_1{R}_2}{R_1+R_2}=\frac{2.5\times 2.5}{2.5+2.5}=1.25\Omega$