Question

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

• A. $60cal/s$
• B. $30cal/s$
• C. $120cal/s$
• D. $100cal/s$

Answer 1

The correct option is C $120cal/s$

Step 1: Calculation of resistance
Let $i_1$ and $i_2$ be the current flowing through upper and lower branch respectively as shown in figure below

Since, the resistance in upper branch is a series combination of $2\Omega$ and $3\Omega$ and resistance in lower branch is a series combination of $6\Omega$ and $4\Omega$, then
Resistance of upper branch $R_1=2+3=5\Omega$
Resistance of lower branch $R_2=4+6=10\Omega$

Step 2:Find ratio of current
Hence, for parallel combination
$\frac{i_1}{i_2}=\frac{R_2}{R_1}=\frac{10}{5}=\frac{2}{1}$
Step 3:Calculation of heat generated
$\frac{\text{heat generated across}3\Omega \left(H_1\right)}{\text{heat generated across}6\Omega \left(H_2\right)}$
$=\frac{(i_1{)}^2\times 3}{(i_2{)}^2\times 6}$
$=\frac{H_1}{H_2}=\frac{4}{2}=2$
heat generated across $6\Omega ,$
$H_2=60cal{s}^{-1}$
heat generated across $3\Omega ,$
$H_1=2H_2=120cal{s}^{-1}$
Therefore heat generated across $3\Omega$ will be $120cal{s}^{-1}$
Hence, option $\left(d\right)$ is correct.