a. The resistivity of copper is 1.72E-8. A copper wire is 15mlong and the diamet

Question

a. The resistivity of copper is 1.72E-8. A copper wire is 15mlong and the diameter is .05cm. What is the resistance of thewire. b. Explain why safe current carrying capacity of an extensioncord is limited by both its length and diameter. c. Commonly used tiny christmas tree lights can operatewith a maximum of 2.5V across their terminals. If the householdvoltage is 120V, esplain how and why the bulbs should be wired.What is the minimum number or bulbs that can be used on the string?What is usually the major frustration of this type of wiring? d. What basic conservation laws are the basis of the Kirchoffrules? a. The resistivity of copper is 1.72E-8. A copper wire is 15mlong and the diameter is .05cm. What is the resistance of thewire. b. Explain why safe current carrying capacity of an extensioncord is limited by both its length and diameter. c. Commonly used tiny christmas tree lights can operatewith a maximum of 2.5V across their terminals. If the householdvoltage is 120V, esplain how and why the bulbs should be wired.What is the minimum number or bulbs that can be used on the string?What is usually the major frustration of this type of wiring? d. What basic conservation laws are the basis of the Kirchoffrules?

Explanation / Answer

a) Resistance,R= l /A ,             Given,=1.72x10-8 m, l=15m and A= (0.05/2)2x10-4 m2           Now, R= 1.72X10-8 x 15/ (0.05/2)2x10-4 = 1.316 b)Current carrying capacity depends on resistance of theconductor, when voltage applied is consatnt.    Now, according to Ohm's Law, I=V/R     As resistance is depending on lendth andcross section area ofthe conductor, obiviously, safe currentcarrying capacity of an extension cord depends on its length anddiameter. c) Maximum safe voltage of the bulb is 2.5V If the supply voltage exceeds this value bulb may fusedoff. As such, number of bulbs are connected in series to avoidthis. Now in the present context, supply voltage is 120V. Minmum number of bulbs thst can be connected in series are120/2.5 = 48 Major frustation of this type of wiring is ,if number of bulbsincreases more than its rated voltage the brightnessdecreases. d) Law of coservation of charge and law of conservation ofenergy                I hope this may help you

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