Explanation Given data: The length of the coaxial resistor is l . The inner radius of coaxial resistor is a . The outer radius of coaxial resistor is b . The conductivity of the material from r = 0 to r = a is σ 1 . The conductivity of the material from r = a to r = b is σ 2 . Calculation: The standard expression for the resistance of the conductor is, R = l σ A ( Ω ) (1) Here, R is the resistance of the conductor. A is the circular cross sectional area of the conductor. σ is the conductivity of the material. The resistance of the inner and outer material of the coaxial resistor is in parallel with each other. So, the total resistance R of the conductor is, 1 R = 1 R inner + 1 R outer (2) Here, R inner is the resistance of the material from r = 0 to r = a . R outer is the resistance of the material from r = a to r = b . Write the standard expression for the area of circular cross section. A = π r 2 (3) Here, r is the radius of the circular cross-section. For area of inner circle A inner , Substitute a for r and A inner for A in the above equation. A inner = π a 2 For area of outer circle A outer , Substitute ( b − a ) for r and A outer for A in the equation (3). A outer = π ( b − a ) 2 For the resistance of inner circle R inner , Substitute π a 2 for A and σ 1 for σ in equation (1)

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Chapter 4, Problem 44P

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The proof that the resistance of the coaxial resistor with its end capped with conducting plate is l[π(σ1a2+σ2(b2−a2))].

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Chapter 4, Problem 43P

Chapter 4, Problem 45P

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The proof that the resistance of the coaxial resistor with its end capped with conducting plate is l [ π ( σ 1 a 2 + σ 2 ( b 2 − a 2 ) ) ] .

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The proof that the resistance of the coaxial resistor with its end capped with conducting plate is l[π(σ1a2+σ2(b2−a2))].

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Chapter 4 Solutions