A rectangular block of copper has sides of length 10 cm, 20cm, and 40 cm. If the block is connected to a 6V sourceacross two of its opposite faces, what are (a) the maximum currentand (b) the minimum current that the block can carry?

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Roosevelt Houghton · Accepted Answer

given length L = 40 cm = 0.40 m breadth B = 20 cm =0.20 m thickness t = 10 cm = 0.10 m Voltage V = 6.0 V area A = (L)(B) we want to find the current we have given voltage (V) and sides of the rectangular box we can proceed with ohms law V=(I)(R) here resistance (R) is unknown from theory when the length and area are given we have the equation for resistance as R = (L/A) where  is the specific resistance of thematerial, (L) is the length, and (A) is the area of the rectangular box as the rectangular box has three sides we havethree values of (L/A) with 0.6 m, 0.22 m, 0.44 m a) from those values of (L/A) choose minimumvalue of (L/A) and substitute in Imax=VRmin =V(LA)max b) from those values of (L/A) choose maximumvalue of (L/A) and substitute in Imin=VRmax =V(LA)max

Jeffrey Jordon · Accepted Answer

Step 1  In this problem, a copper block has dimensions l1=10 cm=0.10 m,l2=20 cm=0.20 m, and l3=40 cm=0.40m  Step 2  From part (a), the possible currents are  I1=2.8×108 A  I2=7.1×107 A  I3=1.8×107 A  Step 3  We see that the minimum is  Imin=1.8×107 A

A rectangular block of copper has sides of length 10 cm, 20cm, and 40 cm. If the block is connected to a 6V sourceacross two of its opposite faces, what are (a) the maximum currentand (b) the minimum current that the block can carry?

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