Answer the following conceptual questions without using formulas to support your

Question

Answer the following conceptual questions without using formulas to support your answer! (a) (5 points) Figure 4 shows cross sections through three wires of identical length and material; the sides are given in millimeters. Rank the wires according to their resistance (measured end to end along each wire's length), greatest first. Figure 4: Question 4a (b) (5 points) Two light bulbs, one 75 W bulb and one 150 W bulb, are connected in parallel with a standard 120 volt ac electrical outlet. The brightness of a light bulb is directly related to the power it dissipates. Therefore, the 150 W bulb appears brighter. How does the brightness of the two bulbs compare when these same bulbs are connected in series with the same outlet? EXPLAIN WHY (recall how resistance is related to power ratings on light bulbs).

Explanation / Answer

a)

resistance R=rho*l/A

here,

rho is resistivity,

l is length

A is cross-section area

wire A:

R1=rho*l/(5*5)*10^-6

R1=rho*l/(25*10^-6) -------(1)

wire B:

R2=rho*l/(2*8)*10^-6

R2=rho*l/(16*10^-6) -------(2)

wire C:

R3=rho*l/(3*6)*10^-6

R3=rho*l/(18*10^-6) -------(3)

from equation (1), (2) and (3)

===> R2>R3>R4

b)

bulb 1 power P1=75 w

bulb 2 power P2=150 w

here,

bulb2 power is greater,

hence bulb2 appears brighter

now,

voltage v=120v

bulbs are in parallel combination,

then,

P1=V^2/R1

75=120^2/R1 ---> resistance of bulb1 is R1=192 ohm

and

P2=V^2/R2

150=120^2/R2 ---> resistance of bulb1 is R2=96 ohm

now,

these two bulbs are connected in series combination,

power at bulb1 is ---> P1=i^2*R1

power at bulb2 is ---> P2=i^2*R2

here

current(i) is same for both bulbs,

then power of bulb1 is greater then the power of bulb2

hence,

hence bulb1 appears brighter

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