Bulb D in the circuit above is then removed from its socket. (What happens when

Question

Bulb D in the circuit above is then removed from its socket. (What happens when a bulb is removed? Study the description of a bulb's construction on page 2 of the Electric Circuits I: Current lab if you are unsure.) When bulb D is removed from its socket,

the brightness of bulb F would:  ---Select--- Increase. Decrease. Stay the same. Not enough information.

the brightness of bulb C would:  ---Select--- Increase. Decrease. Stay the same. Not enough information.

Explain your reasoning for the choices you have made.

Explanation / Answer

Let R be the resistance of each bulb.With Light D in circuit ,the If Bulb D is removed from the circuit ,then equivalent resistance of right branch is

Rright=RA+RC =R+R =2R

equivalent resistance of left branch is

RLeft =RB+RF =R+R =2R

So current throught each branch is

Ileft = Iright =E/2R =(1/2)(E/R)

Current through Bulb F and Bulb C

IC= IF =E/2R =(1/2)(E/R)

.With Light D in circuit ,,then equivalent resistance of right branch is

Rright=RB+RC||(RD+RE) =R+R*(R+R)/(R+R+R) =(5/3)R

equivalent resistance of left branch is

RLeft =RB+RF =R+R =2R

So current throught each branch is

Iright =E/(5/3)R =(3/5)(E/R)

Ileft =E/2R

Current through Bulb C and Bulb F are

IC=E/2R

IF =(3/5)(E/R)(2R/R+2R)=(2/5)(E/R)

So from above calculations

a)

The Brightness of bulb F would increase ,since more current is flowing i.e Ibefore=(2/5)(E/R) is less than Iafter=(1/2)(E/R)

b)

The Brightness of bulb C would stay the same since same current E/2R flows in both the cases

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