   Chapter 17, Problem 20P

Chapter
Section
Textbook Problem

A rectangular block of copper has sides of length 10. cm, 20. cm, and 40. cm. If the block is connected to a 6.0-V source across two of its opposite faces, what are (a) the maximum current and (b) the minimum current the block can carry?

(a)

To determine
The maximum and minimum current carried by the copper block

Explanation

Given Info: The rectangular copper block has the sides 10cm,20cm,40cm The potential difference applied across the opposite sides are 6.0V , The rectangular copper block has the sides 10cm,20cm,40cm . The potential difference applied across the opposite sides are 6.0V .

Explanation:

Formula to calculate the resistance of copper block is

R=ρlA

• R is the resistance of copper bock
• ρ is the resistivity of copper,
• A is the area of cross section of copper block,
• l is the length of the copper block,

Suppose the voltage is applied in opposite faces separated by 10cm , then the area of cross section is 20cm×40cm

Substitute 10cm for l , 1.70×108Ωm for ρ and 20cm×40cm for A in the above equation to find R1

R1=(1.70×108Ωm)(10cm)(20cm×40cm)(1cm10-2m)=2.125×108Ω

Suppose the voltage is applied in opposite faces separated by 20cm , then the area of cross section is 10cm×40cm

Substitute 10cm for l , 1.70×108Ωm for ρ and 20cm×40cm for A in the above equation to find R2

R2=(1.70×108Ωm)(20cm)(10cm×40cm)(1cm10-2m)=8.5×108Ω

Suppose the voltage is applied in opposite faces separated by 40cm , then the area of cross section is 10cm×20cm

Substitute 40cm for l 1

(b)

To determine
The minimum current of copper block

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