Three builm of identical resistance see pland ciresit seshown. It bulb-C, burns

Question

Three builm of identical resistance see pland ciresit seshown. It bulb-C, burns B. ut (is removed from the cireuit) what happto the brightness of bult B Hi brightnms is about power which is about CURRENT hrough the balb. ExtraHint: Bulb A. will change in brightness nn the ccult is different after bulb "C.s removed C. 5 .AO Bulb "B.. is of88 current amot how now. BO) Wle don't know the EMF ?> we cannot say CO Bulb "B.. is unchanged. DO Bulb "Bg brighter EO Bulb "B get dimer I pt BONUS QUESTION: Dos bulb A-in the peevio mer when "C is removed peoblem get hrighter or din- 8.AO Bulb A. gta dimme BO Bulb 'A ents brightes CO Balb "Ais unchanged C. A square loop of wire with dimensions off L xLmoved through a region of constant uniform magnetie feld of magnitude B. The loop b.fted vekety of y and mows along the x axis. The induced en in the loop is gi by BO Depends how long we move it (e t DO 0 I pt The figure above shows the electric eld as a fution of distance from the center of a sphere. Which of the following sticoereet H.AO This is an insulating sphere of chg -20/ BO This is a conduetive sphere of chag CO This is an insulating sphere of charge Q-0/ DO This is a conductive sphere of charge Q-10/ EO This is a conductive sphere of charge Q-20/

Explanation / Answer

1) BECAUSE B AND C ARE IN PARALLEL THE RESISTANCE IS REDUCED TO HALF SO THE POTENTIAL DIFFERENCE ACROOS BC COMBINATION IS LESS THAN THAT ACROSS A BUT WHEN WE REMOVE C VOLTAGE ACROSS B= VOLTAGE ACROSS B SO VOLTAGE ACROSS B INCREASES WHICH INCRESES CURRENT THROUGH B THUS INCRESING THE BRIGHTNESS.

2) AS VOLTAGE ACROSS B INCREASES VOLTAGE ACROSS A DECREASES THUS DECREASING ITS BRIGHTNESS.

3) THERE WILL BE NO INDUCED EMF AS FLUX IS NOT CHANGING ACROSS THE LOOP

4)THIS IS DEFINITELY NOT CONDUCTING OTHERWISE THE FEILD INSIDE IT WOULD HAVE BEEN 0. INSULATING. FEILD AT R=2 E=10=KQ/R2=KQ/4 ==> Q=40/K

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