Engineering Board Exam Reviewer · Series Circuits · medium

In the series circuit shown, V=24 V, R1=33, , R2=27, , and R3=60, . How much power is dissipated by R2?

  1. 21.33 W
  2. 4.8 W
  3. 2.16 W
  4. 1.08 W

Answer

D. 1.08 W

Solution

I=V/RT= 24 V120, =200 mA. Power in R2: P=I2R2=(200 mA)2(27, )=1.08 W.

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