In the figure below, you will attach one or more light bulbs of R 28.12 n in par

Question

In the figure below, you will attach one or more light bulbs of R 28.12 n in parallel with the battery (with E 20 V, internal resistance r a.) For an ideal battery (r-0), what would be power dissipated in the first bulb if it was: connected alone P 14.22 v W connected as one of four bulbs P 14.22 v W (How do these compare? b.) For a non-ideal battery where r 1.1 n, i.) Re-answer part a. Also, in each case find the voltage drop that would be measured across the battery terminals connected alone P 13.174 W Vbatt 17 connected as one of four bulbs P 615 Vbatt 23.56 X V (Compare your P values to part a. Compare your Vbatt values to E. Explain.) ii.) CHALLENGE Suppose you wish to dim the first bulb, such that its power output is at or below of its value when connected alone. Find the minimum number of total bulbs you must connect in parallel to make this happen. solve for a general expression so you can use it for all the cases below. 0% bulbs 35% bulbs 10% bulbs

Explanation / Answer

b)

I=E/(r+R)=20/(1.1+28.12)=0.6845 A

P=I2R=0.68452*28.12=13.174 W

Vbat=E-Ir=20-0.6845*1.1=19.247 Volts

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equivalent resistance of 4 bulbs

1/Req=1/28.12 +1/28.12 +1/28.12 +1/28.12

Req=7.03 ohms

current flowing in the circuit

I=E/(Req+r)=20/(7.03+1.1)=2.46 A

current flowing through each resistor

I1=I2=I3=I4=2.46/4=0.615 A

Power dissipated in one bulb

P=0.6152*28.12=10.63 W

Vbat=20-2.46*1.1=17.3 Volts

c)

case1:50 percent

Power in each bulb

P=Vbat2/R=0.5*14.22=7.11

Vbat=sqrt(7.11*28.12)=14.14 volts

since

I=E-Vbat/r =(20-14.14)/1.1=5.33 A

since

I=E/(r+Req)

5.33=20/(1.1+Req)

Req=2.654

since Req=R/N

N=28.12/2.654 =10.6=11 bulbs

case 2:35 percent

Vbat=sqrt(28.12*0.35*14.22)=11.83 V

I=20-11.83/1.1=7.43 A

=>7.43=20/(1.1+Req)

Req=1.593 ohms

N=28.12/1.593=18 bulbs

case3:10 percent

Vbat=sqrt(28.12*0.1*14.22)=6.62 V

I=20-6.32/1.1=12.43 A

=>12.43=20/(1.1+Req)

Req=0.51 ohms

N=28.12/0.51=55 bulbs

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