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Resistance between two opposite faces of square thin film of area 1 mm2 having thickness of1 μm if resistivity of material is 10 6 Ω will be:A) 1000 Ω C) 1 ΩB) 100 Ω D) 10 ΩQ.18 Total resistance between ‘A’ and ‘B’ in the given circuit is:5 ΩA5 Ω 5 Ω 5 ΩBA) 5.6 Ω C) 0.33 ΩB) 3.33 Ω D) 6.6 ΩA B6 Ω 6 Ω 6 Ω

Question

Resistance between two opposite faces of square thin film of area 1 mm2 having thickness of1 μm if resistivity of material is 10 6 Ω will be:A) 1000 Ω C) 1 ΩB) 100 Ω D) 10 ΩQ.18 Total resistance between ‘A’ and ‘B’ in the given circuit is:5 ΩA5 Ω 5 Ω 5 ΩBA) 5.6 Ω C) 0.33 ΩB) 3.33 Ω D) 6.6 ΩA B6 Ω 6 Ω 6 Ω

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Solution

Para resolver la primera parte de la pregunta sobre la resistencia de una película delgada cuadrada, utilizamos la fórmula de la resistencia:

R=ρLA R = \rho \frac{L}{A}

donde:

  • R R es la resistencia,
  • ρ \rho es la resistividad del material,
  • L L es la longitud del material,
  • A A es el área de la sección transversal.

Dado que la película es cuadrada con un área de 1mm2 1 \, \text{mm}^2 y un grosor de 1μm 1 \, \mu\text{m} :

  1. Convertimos las unidades a metros:

    • Área A=1mm2=1×106m2 A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2
    • Grosor t=1μm=1×106m t = 1 \, \mu\text{m} = 1 \times 10^{-6} \, \text{m}
  2. La longitud L L es el grosor de la película, ( L = 1 \times 10^{-6} , \text

This problem has been solved

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