# 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

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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