In the circuit as shown in the figure, the heat produced by 6 ohm resistance due to current flowing in it is 60 calorie per second. The heat generated across 3 ohm resistance per second will be

  • Option 1)

    30 cal

  • Option 2)

    60 cal

  • Option 3)

    120 cal

  • Option 4)

    100 cal

 

Answers (1)
V Vakul

As we learnt in 

Power dissipiated in external resistance -

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

-

 

 For\ 6 \Omega,    P = 60 \times 4.2 J/s

                                 = 252 \omega= I_{1} ^{2}R

\Rightarrow I ^{2}_1= \frac{252}{6}= 42A

 For\ 3 \Omega,  I_{2}= \left(\frac {10}{5} \right )I_{1}= 2I_{1}

\therefore Rate of Heat production =I_{2}^{2} \times 3 = 4(I_{1}^{2} \times 3)= 12 I_{1}^{2}

= 12 \times 42 J = \frac{12 \times 42}{4.2}cal=120\ cal 

 

 

 


Option 1)

30 cal

Incorrect

Option 2)

60 cal

Incorrect

Option 3)

120 cal

correct

Option 4)

100 cal

Incorrect

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