here,
using KVL for the loop
(16 + 1 + 1) * I = 12 - 3
I = 0.5 A
the power dissapated in 16 ohm resistor , P = I^2 * 16
P = 4 W
the power dissapated in 16 ohm resistor is 4 ohm
The power dissipated in the 16-Ohm resistor in the figure is Selected Answer: 4.0 W Answers:8.0...
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