Current Electricity and Electrical InstrumentHard

Question

For the circuit shown in the figure
     

Options

A.the current I through the battery is 7.5 mA
B.the potential difference across RL is 18 V
C.ratio of powers dissipated in R1 and R2 is 3
D.if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9.

Solution

         
24 - 2 × 103 I - 6 × 103 (I - i) = 0
24 - 2 × 103 I - 1.5 × 103 i = 0
Hence I = 7.5 mA
i = 6mA
         
24 - 6 × 103I′ - 2 × 103 (I′ - i′) = 0
24 - 6 × 103 I′ - 1.5 × 103 i′ = 0
I′ = 3.5 mA
i′ = 2mA
= 9

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