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Q. In the circuit diagram ,heat produces in $R, 2R$ and $1.5 \,R$ are in the ratioPhysics Question Image

KCETKCET 2013Current Electricity

Solution:

We have, $l_{1}=\frac{l \times 2 R}{3 R}=\frac{2l}{3}$
$\therefore H_{1}=l_{1}^{2} R=\frac{4 l^{2}}{9} \times R\,\,\,\,\,\,\,\,...(i)$
Also, $\ I_{2}=\frac{l \times R}{3 R}=\frac{1}{3}$
$\therefore H_{2}=l_{2}^{2}(2 R)=\frac{l^{2}}{9} \times 2 R\,\,\,\,\,\,\,\,...(ii)$
and $ H_{3}=R^{2}(1.5 R)\,\,\,\,\,\,\,\,...(ii)$
From Eqs. (i), (ii) and (iii),
$H_{1}: H_{2}: H_{3} =\frac{4 l^{2}}{9} \times R: \frac{R^{2}}{9} \times 2 R: R^{2} \times 1.5 R $
$=\frac{4}{9}: \frac{2}{9}: 1.5=4: 2: 13.5 $
$=8: 4: 27$