In the circuit shown in figure, the heat produced in the 5  Ω resistor due to the current flowing through it is 10 cals/s.  The heat generated in the 4  Ω resistor is

# In the circuit shown in figure, the heat produced in the $5\text{\hspace{0.17em}\hspace{0.17em}}\Omega$ resistor due to the current flowing through it is 10 cals/s.  The heat generated in the $4\text{\hspace{0.17em}\hspace{0.17em}}\Omega$ resistor is 1. A

1 cal/s

2. B

2 cal/s

3. C

3 cal/s

4. D

4 cal/s

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### Solution:

$\begin{array}{l}{\mathrm{P}}_{5}={\mathrm{i}}_{5}^{2}\mathrm{R},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{P}}_{5}=\left({\mathrm{i}}_{5}^{2}\right)5\\ {\mathrm{i}}_{4}=\frac{{\mathrm{i}}_{5}}{2}\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{P}}_{4}\text{\hspace{0.17em}}{\left(\frac{{\mathrm{i}}_{5}}{2}\right)}^{2}\text{\hspace{0.17em}\hspace{0.17em}}×\text{\hspace{0.17em}\hspace{0.17em}}4\\ \frac{{\mathrm{P}}_{4}}{{\mathrm{P}}_{5}}=\frac{1}{5}\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{P}}_{4}=\frac{{\mathrm{P}}_{5}}{5},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{P}}_{4}=\frac{10}{4}\text{\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{cal}/\mathrm{s}\end{array}$

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