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Four resistors, R1 = 18.1 ohm, R2 = 40.3 ohm, R3 = 95.1 ohm and R4 = 20.0 ohm are connected to a 18.0 V battery as shown in the figure. Determine the power dissipated by resistor R2.

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R1 & R2 in series & R3 in parallel )

whole this in series with R4

so for Req

18.1+40.3 =58.4

now this in parallel with 95.1

so 58.4*95.1 /58.4+95.1 =36.18

this in series with 20

so Req= 56.18 ohm

so I through battrey

18/56.18 =0.320 A

currwnt through R2 =0.32*95.1/18.1+40.3+95.1 =0.198 A

so power through R2 =I^2*R2 =1.58 W (ans)
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