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Question:

For the circuit shown in the figure, the ratio of powers dissipated in R1 and R2 is 3. The current I through the battery is 7.5 mA. The potential difference across RL is 18 V. If R1 and R2 are interchanged, the magnitude of the power dissipated in RL will decrease by a factor of 9.

ratio of powers dissipated inR1andR2is 3

ifR1andR2are interchanged, magnitude of the power dissipated inRLwill decrease by a factor of 9.

the current I through the battery is 7.5 mA

the potential difference acrossRLis 18 V

Solution:

24✕10³I - 5.5✕10³i = 0
Hence I = 7.5mA, i = 6mA
24✕10³I' - 5.5✕10³i' = 0
I'=3.5mA, i'=2mA
P₁/P₂ = 6²/2² = 9