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Three equal resistance's each of R ohm are connected as shown in figure. A battery of 2 V and internal resistance 0.1\Omega is connected across the circuit. Calculate the value of R for which the heat generated in the circuit is maximum.

 

 

Option: 1

0.2\Omega 


Option: 2

0.3\Omega 


Option: 3

0.4\Omega


Option: 4

0.5\Omega


Answers (1)

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With respect to points A and B, the three resistances are connected in parallel as shown in figure. The equivalent resistance is given by

\frac{1}{R^{\prime}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R} \quad \therefore R^{\prime}=R / 3

Current i flowing in the circuit

i=\frac{E}{R^{\prime}+r}=\frac{2}{R / 3+0.1}

Heat produced

H=i^2 R^{\prime}=\frac{4 R^{\prime}}{[(R / 3)+0.1]^2}=\frac{4 R}{3[(R / 3)+0.1]^2}

Heat generated in the circuit is maximum when \frac{dH}{dR}==0

Applying this condition we get

R=0.3\Omega

 

 

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