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Q.

For the circuit shown in the figure

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a

the current I through the battery is 7.5 mA

b

the potential difference across RL is 18 V

c

if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9

d

ratio of powers dissipated in R1and R2 is 3

answer is A, D.

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Detailed Solution

Total resistance of the circuit R=2+6x1.56+1.5=4.2kΩ

Now I=244.2x103=7.5x10-3A

Given RLi1=7.5x10-3x6(6+1.5)=6x10-3A

Now Current in RL=3.5x10-3x2(2+1.5)=2x10-3A

Power ,P1'=(2x10-3)2x1.5x103=6x10-3 J 

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