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### IIT-JEE 2010 Paper 1 Offline

Numerical

When two identical batteries of internal resistance 1 $$\Omega$$ each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25 J2, then the value of R in $$\Omega$$ is __________.

Correct Answer is 4

## Explanation

In series : When the batteries are connected in series, we have

$${J_1} = {\left( {{{2E} \over {R + 2}}} \right)^2}R$$ In parallel : When the batteries are connected in parallel, we have

$${J_2} = {\left( {{E \over {R + (1/2)}}} \right)^2}R$$ It is given that,

$${{{J_1}} \over {{J_2}}} = 2.25$$

$$\Rightarrow {4 \over {{{(R + 2)}^2}}} \times {{{{(2R + 1)}^2}} \over 4} = 2.25 \Rightarrow {{2R + 1} \over {R + 2}} = 1.5$$

$$\Rightarrow 2R + 1 = 1.5R + 3 \Rightarrow 0.5R = 2$$

Therefore, $$R = 4\Omega$$.

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