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Zorluk: OrtaDirect, Inverse, Joint and Partial Variation

The electrical resistance RR of a wire varies directly as its length LL and inversely as the square of its diameter dd. If a wire of length 36 m36\text{ m} and diameter 3 mm3\text{ mm} has a resistance of 16 Ω16\ \Omega, what is the resistance, in ohms, of a wire of the same material with a length of 45 m45\text{ m} and a diameter of 5 mm5\text{ mm}?

Cevap: 7.2 ohms

Cevap

The resistance of the wire is 7.2 ohms.
The equation governing the relation is R=kLd2R = \frac{kL}{d^2}. Substituting the initial parameters R=16 ΩR=16\ \Omega, L=36 mL=36\text{ m}, and d=3 mmd=3\text{ mm} gives 16=36k9=4k16 = \frac{36k}{9} = 4k, which yields k=4k = 4. Using k=4k = 4 with the new dimensions L=45 mL=45\text{ m} and d=5 mmd=5\text{ mm} gives R=4×4552=18025=7.2 ΩR = \frac{4 \times 45}{5^2} = \frac{180}{25} = 7.2\ \Omega.

Adım Adım Çözüm

1
Formulate the joint variation equation
R=kLd2R = \frac{kL}{d^2}
Direct variation places length LL in the numerator and inverse variation of the square of diameter dd places d2d^2 in the denominator.
2
Determine the variation constant kk
k=4k = 4
Substituting R=16R = 16, L=36L = 36, and d=3d = 3 gives 16=36k9    16=4k    k=416 = \frac{36k}{9} \implies 16 = 4k \implies k = 4.
3
Calculate the new resistance
R=7.2 ΩR = 7.2\ \Omega
Substituting k=4k = 4, L=45L = 45, and d=5d = 5 into R=kLd2R = \frac{kL}{d^2} yields R=4×4525=7.2R = \frac{4 \times 45}{25} = 7.2.

Anahtar Kavram

Joint Variation involving direct proportionality and inverse square law
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