First slide

combination of resistance

Question

A 12 cm wire is given a shape of a right angled triangle A B C having sides 3 cm, 4 cm and 5 cm as shown in the figure. The resistance between two ends (A B, BC, CA) of the respective sides are measured one by one by a multi-meter. The resistances will be in the ratio

Moderate

A

$9 : 16 : 25$
B

$27 : 32 : 35$
C

$21 : 24 : 25$
D

$3 : 4 : 5$

Solution

Let $ρ$ and A be resistivity and area of cross-section of the wire respectively. The wire is bent in the form of right triangle as shown in adjacent figure.
$Resistance of side A B is R_{1} = \frac{ρ 3}{A} Resistance of side B C is R_{2} = \frac{ρ 4}{A} Resistance of side A C is R_{3} = \frac{ρ 5}{A} ∴ The resistance between the ends A and B is R_{A B} = \frac{R_{1} (R_{2} + R_{3})}{R_{1} + R_{2} + R_{3}} = \frac{\frac{3 ρ}{A} (\frac{4 ρ}{A} + \frac{5 ρ}{A})}{\frac{3 ρ}{A} + (\frac{4 ρ}{A} + \frac{5 ρ}{A})} = \frac{27}{12} \frac{ρ}{A} The resistance between the ends B and C is R_{B C} = \frac{R_{2} (R_{1} + R_{3})}{R_{2} + R_{1} + R_{3}} = \frac{\frac{4 ρ}{A} (\frac{3 ρ}{A} + \frac{5 ρ}{A})}{\frac{4 ρ}{A} + \frac{3 ρ}{A} + \frac{5 ρ}{A}} = \frac{32}{12} \frac{ρ}{A} The resistance between the ends A and C is R_{A C} = \frac{R_{3} (R_{1} + R_{2})}{R_{3} + R_{1} + R_{2}} = \frac{\frac{5 ρ}{A} (\frac{3 ρ}{A} + \frac{4 ρ}{A})}{\frac{5 ρ}{A} + \frac{3 ρ}{A} + \frac{4 ρ}{A}} = \frac{35}{12} \frac{ρ}{A} ∴ R_{A B} : R_{B C} : R_{A C} = \frac{27}{12} : \frac{32}{12} : \frac{35}{12} = 27 : 32 : 35$

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