In the circuit analysis we\'ve done, we assumed that the resistance of the wires

Question

In the circuit analysis we've done, we assumed that the resistance of the wires is negligible compared to that of the circuit elements. This is usually true, but under some circumstances even that small resistance can become significant. Consider the 14-gauge copper wire used for typical household applications. It has a cross sectional area of 2.08 times 10^-6 m^2, and copper leas a resistivity of 1.72 times 10^-8 ohm m. A particular device uses 176 in of this wire exposed to the maximum current. Find the resistance of this length of wire. If this wire feeds a "load" (which can be a very complicated circuit) which has a resistance of 3.22 ohm, running on an ordinary 120 V household voltage, how much current will the whole system draw? Remember dial the wire resistance will be significant in this case, and must be added. Find the power consumed by the whole circuit. How much of (hat power will be used (dissipated, wasted) just in the wires?

Explanation / Answer

Resistance = R = rho*L/A

L = length of the wire

A = area of cross section

rho resistivity

Resistance = R = (1.72*10^-8*176)/(2.08*10^-6) = 1.46 ohms

__________________

(b)

total ressitace of the circuit = 3.22 + 1.46 = 4.68 ohms

current = i = E/Rtot = 120/4.68 = 25.6 A

(c)

P = i^2*Rtot = 25.6^2*4.68 = 3067.1 W

(d)

power in wire = i^2*R = 25.6^2*1.46 = 956.8 W

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