how many bulbs of resistance 6 ohm should be joined in parallel to draw a current of 2 amperes from a battery of 3 volts.?

Solution:

Suppose n resistances of 6 ohm are connected in parallel and the resultant resistance=R

Then,

1/R=n/6=>R=6/n

Now Current I=2 A,Voltage V=3 volts

By ohm's law V=IR

3=2x6/n

n=4

Hence 4 bulbs are connected in parallel. 

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V=3V

I=2A

R =V/I=3/2            

R  OF I BULB=6 ohms

1/R(SUM OF ALL R's)=1/R1+1/R2+1/R3.....1/Rn

LET THE NO.OF BULBS BE x

1/R=X/6

3/2=X/6

9 BULBS=X

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