Get Answers to all your Questions

header-bg qa

 Two electric bulbs, rated at (25 W, 220 V) and (100 W,220 V ) are connected in seriees across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then :

  • Option 1)

     P1 = 16 W , P2 = 4 W

  • Option 2)

     P1 = 4 W , P2 = 16 W

  • Option 3)

     P1 = 16 W , P2 = 9 W

  • Option 4)

     P1 = 9 W , P2 = 16 W

Answers (1)

best_answer

 

Power dissipiated in external resistance -

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

-

B_{1}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, B_{2}

25W_{1}220V\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \ 100W_{1}220V

R_{1}=\frac{220^{2}}{25}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \ R_{2}=\frac{220^{2}}{100}

R_{1} & R_{2}  are in series

R_{nct}=R_{1}+R_{2}

I_{nct}=220V

I_{nct}=\frac{220}{R_{1}+R_{2}}

P_{1}=I_{nct}^{2}{R_{1}

P_{2}=I_{nct}^{2}{R_{2}

P_{1}=\left ( \frac{220}{R_{1}+R_{2}} \right )^{2}\times R_{1}

        = \frac{220\times 200}{\left ( R_{1}+R_{2} \right )^{2}}\times R_{1}

         = \left ( \frac{220}{220\left ( 220 \right )\left ( \frac{1}{25} +\frac{1}{100}\right )} \right )^{2}\times \frac{220\times 220}{25}

           =\left ( \frac{1\times 25\times 100}{220\times125} \right )^{2}\times\frac{\left ( 220 \right )^{2}}{\left ( 25 \right )}

             =\left ( \frac{100}{125} \right )^{2}\times 25

              =\left ( \frac{4}{5} \right )^{2}\times 25

                   P_{1}=16W

               P_{2}=P_{1}\times \left ( \frac{R_{2}}{R_{1}} \right )

                      =16\times \left ( \frac{25}{100} \right )

                      P_{2}=4W

 

 


Option 1)

 P1 = 16 W , P2 = 4 W

Option 2)

 P1 = 4 W , P2 = 16 W

Option 3)

 P1 = 16 W , P2 = 9 W

Option 4)

 P1 = 9 W , P2 = 16 W

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE