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  • Solve it, The figure shows three circuits I, II and III which are connected to a 3V battery. If the powers dissipated by the configurations I, II and III are P1, P2 and P3 respectively, then :

The figure shows three circuits I, II and III which are connected to a 3V battery.  If the powers dissipated by the configurations I, II and III are P1, P2 and P3 respectively, then :

 

  • Option 1)

    P1 > P2 > P3

  • Option 2)

    P1 > P3 > P2

  • Option 3)

     P2 > P1 > P3

  • Option 4)

     P3 > P2 > P1

     

 

Answers (2)

best_answer

As we learnt in

Power dissipiated in the circuit -

P= \left (\frac{E}{R+\frac{r}{n}} \right )^{2}\cdot R         

-

 

  R_{1}=1\Omega \: R_{11}=\frac{1}{2\Omega }\: R_{111}=\frac{3}{2}\OmegaP=\frac{V^{2}}{R}= P\propto \frac{1}{R}

R_{111}> R_{1}> R_{11}

Hence P_{3}< P_{1}< P_{2}

Power (P)=P=\frac{V^{2}}{R}= P\propto \frac{1}{R}

 

  


Option 1)

P1 > P2 > P3

This option is incorrect  

Option 2)

P1 > P3 > P2

This option is incorrect 

Option 3)

 P2 > P1 > P3

This option is correct 

Option 4)

 P3 > P2 > P1

 

This option is incorrect 

Posted by

Aadil

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