The circuit shown below has 4 light bulbs connected to anideal battery. The resi

Question

The circuit shown below has 4 light bulbs connected to anideal battery. The resistances of the bulbs are: RA = RD = R and R B = R C = 2R a) rank the relative brightness of these bulbs.Explain your rankings using the appropriate physics principle(s).DO NOT USE NUMBERS (brightness is proportional tothe power dissipated by the bulb). b) Write an expression for the current (expressedin terms of resistance R and potential V) througheach bulb. c) give a qualitative explanation of what happensto the brightness of A, B and C when D is unscrewed. d) An ideal conducting wire is added to thecircuit as shown in figure (b) below. Give a qualitativeexplanation of what happens to the brightness of each bulb when thewire is added.

Explanation / Answer

Since P=I2R in therms of equivalent resistance Re  is Re = RA + ( RB +RC ) || RD Re= R + ( 2R + 2R) R /(2R + 2R + R) Re=R + (4/5)R=(9/5)R The current through resistance RA is IA = V/Re IA = V/((9/5)R)= 5V/(9R) IB = IC = [V - IA RA ]/ ( RB + RC ) IB = IC = [V - [5V/(9R)] R ] / ( 2R +2R ) IB = IC = V[1 - [5 / 9 ] ] / ( 4R   ) IB = IC = V /(9 R ) ID = [V - IA RA ] / RD ID = [V - [5V/(9R)] R ] / R ID = 4V/( 9R) Now we can apply P=I2R For RA PA=IA2RA PA=( 5V/(9R))2 R =PA=( 25V2/(81R)) For RB PB=IB2RB PB=(V /(9 R ))2 2R PB=2 V2 /(81 R ) PB=PC For RDPD=ID2 RD PD=(4V/( 9R))2RPD=16V2/(81R) a)PC = PB

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