In the circuit shown below, find the currents I1, I2, and I3. Also, find the powers on R1, R2, R3, and Vs and state whether power is absorbed or released by each element.
Application of Ohm's law yields
p1 = Vs×I1 = 5 V × 5 mA = 25 mW (absorbed)
p2 = Vs×I2 = 5 V × 2.5 mA = 12.5 mW (absorbed)
p3 = Vs×I3 = 5 V × 1 mA = 5 mW (absorbed)
pVs = Vs×(-I1 - I2 - I3) = -5 V × 8.5 mA = -42.5 mW (released)
p1 + p2 + p3 + pVs = 0
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