Question

The maximum allowed power dissipation for a 27.3-ω resistor is stated to be 10.0 w. find the largest current that this resistor can take safely without burning out.

Answer 1

The power dissipated on a resistor is related to the current I flowing through it and its resistance R by the relationship
$P= I^2 R$
If the resistance is $R=27.3 \Omega$ and the maximum dissipated power is 10.0 W, then we can find the maximum allowed current by re-arranging the previous equation:
$I= \sqrt{ \frac{P}{R} }= \sqrt{ \frac{10.0 W}{27.3 \Omega} }=0.6 A$