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  • I have a doubt, kindly clarify. - Current Electricity - NEET-2

A current of 2A flows through a 2\Omega resistor when connected across a battery. The same batterysupplies a current of 0.5 A when connected across a 9\Omega resistor. The internal resistance of the battery is

  • Option 1)

    0.5\Omega

  • Option 2)

    1/3\Omega

  • Option 3)

    1/4\Omega

  • Option 4)

    1\Omega

 

Answers (1)

As we learnt in

Current given by the cell -

i=frac{E}{R+r}

- wherein

R- External resistance

r-  internal resistance

 

 Let internal resistance is r

I=\frac{E}{r+R} => 2=\frac{E}{r+2}------------(1)and 0.5=\frac{E}{r+9}------------(2)

Divide (1) and (2) \frac{2}{0.5} =\frac{r+9}{r+2} or 4(r+2)=r+9

=>4r+8=r+9=>r=\frac{1}{3}\Omega

 


Option 1)

0.5\Omega

This is incorrect option 

Option 2)

1/3\Omega

This is  correct option 

Option 3)

1/4\Omega

This is incorrect option 

Option 4)

1\Omega

This is incorrect option 

Posted by

subam

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