#1 Four copper wires of equal length are connected in series.Their cross-section

Question

#1 Four copper wires of equal length are connected in series.Their cross-sectional areas are 1 cm2, 2.4 cm2, 4.8 cm2, and 5 cm2.If a voltage of 74 V is applied to the arrangement, determine thevoltage across the 2.4 cm2 wire. Answer in units of V.

#2
a. A string of 22 identical Christmas tree lights are connected inseries to a 125 V source. The
string dissipates 55 W. What is the equivalent resistance of thelight string? Answer in units of Ohm.
b. What is the resistance of a single light? Answer in units ofOhm
c.How much power is dissipated in a single light? Answer in unitsof W.
d. One of the bulbs quits burning. The string has a wire thatshorts out the bulb filament
when it quits burning, dropping the resistance of that bulb tozero. All the rest of the bulbs
remain burning. What is the resistance of the light string now?Answer in units of Ohm
e.How much power is dissipated by the string now? Answer in unitsof W.
.

Explanation / Answer

(Q1) Resistance, R = L / A    ( where = resistivity, L = length and A = crosssectional area ) Here and L are constant R 1/A we know thar V = I R In a series circuit, I is constant V R Hence, V 1/ A ratio of V = ratio of 1/A = 1 : 1/2.4 : 1/4.8 :1/5 = 57.6 : 24 : 12 : 11.52 voltage across 2.4 cm^2 wire = ( 24 / 105.12) 74 = 16.9volts (Q2)         (a)      R = V 2 /P = (125) 2 / 55 =284         (b)      Resistance of single bulb, r= R /22 = 284 / 22 = 12.91          (c)     Voltage on each bulb, v = 125 /22 = 5.68 volts                   Power disspated in a single bulb, p = v 2 / r = ( 5.68 )2 / 12.91 = 2.5 W          (d)     New resistance of string, Rn = (21)(12.91) = 271.1            (e)    Newpower of string , Pn = ( V ) 2 / Rn = (125)2 / 271.1 = 57.64 W

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