F.2 IS - Electric current

1. Two bulbs are connected in series in the following diagram

Diagram : http://i26.photobucket.com/albums/c110/JM2711/Dec%...

If a thick copper wire is connected across A

a.What will happen to bulb B?

b.What will happen to bulb A?

c.What will happen to the reading of the ammeter?

2.Ben has connected the circult as shown in the following diagram

Diagram : http://i26.photobucket.com/albums/c110/JM2711/Dec%...

a.i.What happens when switch S is closed?

ii.What happens when switch S is opened?

b.i.Ben thinks that the circult is properly connected because the switch can be used to switch on or off the bulb.Do you agree?

Answer:Yes

b.ii.Explain your answer in bi. briefly.

Please answer and explain in English.

3 Answers

Rating
  • 天同
    Lv 7
    1 decade ago
    Favorite Answer

    1.a.What will happen to bulb B?

    Since the copper wire by-passed bulb A, leaving only bulb B in the circuit. Bulb B will glow brighter.

    b.What will happen to bulb A?

    Bulb A will not glow. Iy has been by-passed by the copper wire. There is no current flowing through bulb A.

    c.What will happen to the reading of the ammeter?

    The reading will increase showing an increase of current.

    2. (a)(i) What happens when switch S is closed?

    The bulb will not glow, and the battery gets hot.

    ii.What happens when switch S is opened?

    The bulb glows.

    Ben thinks that the circult is properly connected because the switch can be used to switch on or off the bulb.Do you agree?

    No. Closing the switch causes a "short circuit". There is still current flowing through. Since the resistance of the circuit is low, the current flowing will be large, battery might be damaged.

  • 1 decade ago

    (a)

    Since The Bulb A Was Short-Circuit,So The Current Will Be Increase,SO The Bulb B Will More Brightless.

    (b)

    According To Ohms Law V=IR,Copper Wire Cause The Bulb A Short-Circuit,So the Bulb A Will Not Light Up(Very Very Little Amount To Bulb A)

    (c)

    Since The Copper Connect In Parallel,Than According To The Parallel-Resistance Law,1/R=1/R1+1/R2...+1/Rn,

    So The Current Will Increase,When Combine With Bulb B In Series,

    Since The Resistance Is Lower,So The Current Will Increase,Thus,The Ammeter Will Increase The Reading.

    (2a.i.)

    When The Switch Is Closed,Its Means The Two Poles Will Have Continuity To Let The Current Flows.So Therefore,After Closed The Switch,The Resistances Will Be Very Low,So That It Cause Short-Circuit.

    (2a.ii.)

    As The Photo,Switch Is Open,The Resistance Of The Switch Close To Infinity,So Therefore No Current Flow There,So The Bulb Will Light Up.

    (2b.i.)

    THE ANSWER IS YES?!?!

    IT MUST WRONG.

    Although The Switch Can Be Controlled The Bulb On And Off,But Will The Bulb Off,Its A Short-Circuit Actually,It Causes The Batteries Very Hot,Even Explode,The Wire Will Get Very Hot As Well.

    The Switch Can Control The Bulb On And Off,But This Way Is UNSAFE.

  • 1 decade ago

    1a.bulb b will increase brightness

    1b.bulb a will be dimmed or extinguished

    1c.ammeter reading will be increased

    2ai.same as 1b

    2aii.same as 1a or it is light up normally

    2bi.of course not, when switch closed the battery is short circuited which mean the loading is high and the current flow is also high, the power is lost by heating up the copper wire, if in case of household electricty, it will cause fire and damaging.

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