Answer to Question #273851 in Electric Circuits for mickii

Question

The maximum safe current that can pass through a resistor of 2 kΩ rated at 0.25 Watts is

Answer 1

Solution;

$P=I^2R$

Given;

$R=2000\Omega$

$P=0.25$

Therefore;

$I^2=\frac PR$

$I=\sqrt{\frac PR}=\sqrt{\frac{0.25}{2000}}=12.5mA$