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Chapter 24: Problem 57

Two cylindrical resistors are made from the same material and have the same length. When connected across the same battery, one dissipates twice as much power as the other. How do their diameters compare?

Short Answer

Expert verified
The diameter of the first resistor, which dissipates twice as much power, is 鈭�2 times the diameter of the second resistor.

Step by step solution

01

Understanding the Law of Resistance

The resistance of a wire resistor can be written as \( R = 蟻L/A \), where \( 蟻 \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. Since the resistors are cylindrical, their cross-sectional area \( A \) is related to their diameter \( d \) by \( A = 蟺(d/2)^2 \). Since the two resistors have the same length and are made of the same material, they have the same resistivity \( 蟻 \). Therefore, the resistance \( R \) is inversely proportional to the square of the diameter \( d \).
02

Apply the Power Dissipation formula

The power dissipation in a resistor is given by the formula \( P = I^2R \). Since they are connected across the same battery, the current\( I \) is assumed to be the same in both resistors. Thus, the power dissipated \( P \) is directly proportional to the resistance \( R \). We know that \( P1 = I^2R1 \) and \( P2 = I^2R2 \). Since \( P2 = 2P1 \), this simplifies to \( R2 = 2R1 \).
03

Compare their diameters

Through step 1, we expressed \( R \) in terms of diameter \( d \), as \( R 鈭� 1/d^2 \). We have already established that \( R2 = 2R1 \). Substituting the relation of resistance and diameter into this equation gives \( 1/d2^2 = 2(1/d1^2) \). By manipulating this equation, we obtain \( d1 = 鈭�2d2 \). Hence, the diameter of the first resistor is 鈭�2 times the diameter of the second resistor.

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