Question
A rectangular block of copper has sides of length 8 cm, 16 cm, and 33 cm. If the block is connected to a 7.0 V source across two of its opposite faces of the rectangular block, what are the currents that the block can carry?
(a) maximum current
(b) minimum current
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PLEASE do not post this answer:
resistivity of copper = 1.72x8^-8 ohm-m * 100cm/m =
1.72x8^-7 ohm=cm
Rmax = 1.72x8^-7*33/8*16 = 5.41305542 × 10-5
Rmin = 1.72x8^-7*33/33*16 = 1.31225586 × 10-5
(a) Imax = 7/ Rmin = 2.79x8^8 A
(b) Imin = 7 / Rmax = 1.74x8^7 A
i tried these answers and they are incorrect. :[
thank you.
Explanation / Answer
Current I = Voltage(V) / Resistance (R) Clearly for maximum current, theresistance should be minimum and vice-versa. R = * L / A i.e. for higher resistance R area should beminimum and vice-versa. For copper resistivity = 1.7* 10-8 -m a. For minimum resistance, Amax = 16cm * 33 cm = 0.16* 0.33 = 5.28* 10-2 m2 length L = 8cm = 0.08 m Rmin = 1.7* 10-8 * 0.08 / 5.28 * 10-2 = 2.575* 10-8 Current Imax = V/ Rmin = 7.0/ 2.575 * 10-8 = 2.717* 108 A b. For maximumresistance, Amin = 8cm * 16 cm = 0.08* 0.16 = 1.28* 10-2 m2 length L = 33cm = 0.33 m Rmin = 1.7* 10-8 * 0.33 / 1.28 * 10-2 = 4.382* 10-7 Current Imin = V/ Rmax = 7.0/ 3.49 * 10-7 = 1.597* 107 A Current I = Voltage(V) / Resistance (R) Clearly for maximum current, theresistance should be minimum and vice-versa. R = * L / A i.e. for higher resistance R area should beminimum and vice-versa. For copper resistivity = 1.7* 10-8 -m a. For minimum resistance, Amax = 16cm * 33 cm = 0.16* 0.33 = 5.28* 10-2 m2 length L = 8cm = 0.08 m Rmin = 1.7* 10-8 * 0.08 / 5.28 * 10-2 = 2.575* 10-8 Current Imax = V/ Rmin = 7.0/ 2.575 * 10-8 = 2.717* 108 A = 2.717* 108 A = 0.08* 0.16 = 1.28* 10-2 m2 length L = 33cm = 0.33 m Rmin = 1.7* 10-8 * 0.33 / 1.28 * 10-2 = 4.382* 10-7 Current Imin = V/ Rmax = 7.0/ 3.49 * 10-7 = 1.597* 107 A = 1.597* 107 A