Can you help me with 6) b and c please What is the voltage across the A1 wire? 4

Question

Can you help me with 6) b and c please

What is the voltage across the A1 wire? 4.35 times 10^1 V Keeping the same total length. What should be the length of the Cu wire so that each wire segment, Cu and Al, have the same potential difference? 1.74 times 10^1 m A hair dryer consumers P = 1800 Watts of power from a 120 V wall plug. What current does the hairdryer draw? 1.500 times 10^1 A The current is carried from the breaker box to the power outlet by a 14-gauge copper wire, with diameter D1 = 1.628 mm. The wire runs a distance L = 20.0 m from the box to the outlet and the same distance back. The resistivity of copper is p = 1.680 times 10^-8 Ohm m. How much power is consumed due to the resistance of the wire? Suppose the above ratio of power consumed by devices to power consumed by wires is typical in the house. If the house spends $1, 100 per year in electricity bills, how much money can be saved per year by rewiring the house with 12 gauge wire, with diameter D2 = 2.053 mm? (Give your answer in dollars, but do not write "$") A 500 Ohm resistor is connected to a 9 V battery. Calculate the current through this resistor. 1.80 times 10^-2

Explanation / Answer

b) current in the circuit , I = 15 A

resitance of the wire , R = p * L/area

R = 2 * 1.68 *10^-8 * 20/(pi * (1.628 *10^-3/2)^2)

R = 0.323 Ohm

power consummed due to resistance of wire = I^2 * R

power consummed due to resistance of wire = 0.323 * 15^2

power consummed due to resistance of wire = 72.8 W

c) if the wires are used at 2.053 Ohm

resistance , R2 = 2 * 1.68 *10^-8 * 20/(pi * (2.053 *10^-3/2)^2)

R2 = 0.203 Ohm

for 1000 $ bill

money wasted due to resitance = 1000 * 72.8/1800 = 40.44 $

NOw, for the money saved when the new wire is used

money saved = 40.44 * (1 - 0.203/0.323)

money saved = 15.02 $

the money saved is 15.02 $

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