Verified Solution • 2025–26 Session moderate

QJEEM12Ph579f: JEE Main 2012 Physics - Class 12 JEEM12Ph579f Solution: Two electric bulbs marked and are connected in series to a…

Two electric bulbs marked 25W25W −- 220220 VV and 100W100W −- 220V220V are connected in series to a 440440 VV supply. Which of the bulbs will fuse?

Multiple Choice Options

  • A Both
  • B 100100 WW
  • C 2525 WW
  • D Neither

Step-by-Step Verified Solution

The current upto which bulb of marked 25W25W - 220V,220V, will

not fuse I1=W1V1=25220Amp{I_1} = {{{W_1}} \over {{V_1}}} = {{25} \over {220}}Amp

Similarly, I2=W2V2=100220 Amp{I_2} = {{{W_2}} \over {{V_2}}} = {{100} \over {220}}\,Amp

The current flowing through the circuit

AIEEE 2012 Physics - Current Electricity Question 307 English Explanation

I=440Reff,  Reff=R1+R2I = {{440} \over {{{\mathop{\rm R}\nolimits} _{eff}}}},\,\,{{\mathop{\rm R}\nolimits} _{eff}} = {R_1} + {R_2}

R1=V12P1=(220)225;   {R_1} = {{V_1^2} \over {{P_1}}} = {{{{\left( {220} \right)}^2}} \over {25}};\,\,\,

R2=V22P=(220)2100{R_2} = {{V_2^2} \over P} = {{{{\left( {220} \right)}^2}} \over {100}}

I=440(220)225+(220)2100I = {{440} \over {{{{{\left( {220} \right)}^2}} \over {25}} + {{{{\left( {220} \right)}^2}} \over {100}}}}

=440(220)2[125+1100]= {{440} \over {{{\left( {220} \right)}^2}\left[ {{1 \over {25}} + {1 \over {100}}} \right]}}

I=40220  AmpI = {{40} \over {220}}\,\,Amp

as {I_1}\left( { = {{25} \over {220}}A} \right) < I\left( { = {{40} \over {220}}A} \right) < {I_2}\left( { = {{100} \over {200}}A} \right)

Thus the bulb marked 25W25W - 220220 will fuse.

Related Learning Resources & Exam Preparation