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# Confused! kindly explain, 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

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

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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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