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Question

Physics Question on Resistance

A letter ' AA ' is constructed of a uniform wire with resistance 1.0Ω1.0\, \Omega per cmcm, The sides of the letter are 20cm20\, cm and the cross piece in the middle is 10cm10 \,cm long. The apex angle is 6060 . The resistance between the ends of the legs is close to:

A

50.0Ω50.0\,\Omega

B

10Ω10 \,\Omega

C

36.7Ω36.7\,\Omega

D

26.7Ω26.7 \,\Omega

Answer

26.7Ω26.7 \,\Omega

Explanation

Solution

The circuits are shown above.
Using trignometry, 5x=sin30=0.5\frac{5}{ x }=\sin 30=0.5
x=10cm\Rightarrow x =10\, cm
Thus we get AB=CD=2010=10cmA B=C D=20-10=10\, cm
Since all parts are of same length, thus resistance of each part
R=10×1=10ΩR=10 \times 1=10 \,\Omega
Equivalent circuit is also shown above.
Equivalence resistance between BB and EE,
RBE=(10+10)10=2010R _{ BE }=(10+10)\|10=20\| 10
Req=20×1020+10=6.7Ω\therefore R_{ eq }=\frac{20 \times 10}{20+10}=6.7\, \Omega
Equivalent resistance between AA and DD,
RAD=10+6.7+10=26.7ΩR _{ AD }=10+6.7+10=26.7\, \Omega