For the circuit shown in the figure

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the current I through the battery is 7.5 mA

the potential difference across $R_{L}$ is 18 V

ratio of powers dissipated in $R_{1}$ and $R_{2}$ is 3

if $R_{1}$ and $R_{2}$ are interchanged, magnitude of the power dissipated in $R_{L}$ will decrease by a factor of 9

answer is A, D.

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Detailed Solution

Total resistance of the circuit $R = 2 + \frac{6 x 1.5}{6 + 1.5} = 4.2 k Ω$

Now $I = \frac{24}{4.2 x 10^{3}} = 7.5 x 10^{- 3} A$

Given $R_{L}$ , $i_{1} = \frac{7.5 x 10^{- 3} x 6}{(6 + 1.5)} = 6 x 10^{- 3} A$

Now Current in $R_{L} = \frac{3.5 x 10^{- 3} x 2}{(2 + 1.5)} = 2 x 10^{- 3} A$

Power , $P_{1}^{'} = {(2 x 10^{- 3})}^{2} x 1.5 x 10^{3} = 6 x 10^{- 3} J$

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