The resistance of the series combination of two resistances is  S   . When they are joined in parallel the total resistance is   P.    if    S =  nP  then the minimum possible value of  n is .

  • Option 1)

    4

  • Option 2)

    3

  • Option 3)

    2

  • Option 4)

    1

 

Answers (2)

As we learnt in

Power dissipiated in external resistance -

P=(\frac{E}{R+r})^{2}R

-

 

In series combination, S=(R_{1}+R_{2})

In parallel combination, P=\frac{R_{1}R_{2}}{R_{1}+R_{2}}

\because S=nP

\Rightarrow (R_{1}+R_{2})=\frac{nR_{1}R_{2}}{(R_{1}+R_{2})}

\Rightarrow (R_{1}+R_{2})^{2}=nR_{1}R_{2}

for minimum value R_{1}=R_{2}=R

\therefore (R_{1}+R_{2})^{2}=n(R\times R)

=4R^{2}=nR^{2}=n=4


Option 1)

4

This option is correct.

Option 2)

3

This option is incorrect.

Option 3)

2

This option is incorrect.

Option 4)

1

This option is incorrect.

N neha

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