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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.

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skyluke89
The power dissipated on a resistor is related to the current I flowing through it and its resistance R by the relationship
[tex]P= I^2 R [/tex]
If the resistance is [tex]R=27.3 \Omega[/tex] and the maximum dissipated power is 10.0 W, then we can find the maximum allowed current by re-arranging the previous equation:
[tex]I= \sqrt{ \frac{P}{R} }= \sqrt{ \frac{10.0 W}{27.3 \Omega} }=0.6 A [/tex]

The amount of flow of charge per unit time is known as the current. The largest current that this resistor can take safely without burning out will be 0.6 amperes.

What is power dissipation in a resistor?

The process of losing power in the form of heat as a result of the main activity is known as power dissipation. Dissipation of power is a natural occurrence.

All of the circuit's resistors that have a voltage drop across them will dissipate power. Due to the conversion of electrical energy to thermal energy, all resistors will have a power rating.

The power dissipated on a resistor is given as

P = I²R

I is current  flowing through it  

R is the resistance

[tex]\rm {P = I^2R}\\\\I=\sqrt{\frac{P}{R} }\\\\I=\sqrt{\frac{10}{27.36\\}[/tex]

I = 0.6 A

Hence the largest current that this resistor can take safely without burning out will be 0.6 amperes.

To learn more about the power dissipated on a resistor refer to the link;

https://brainly.com/question/10728846

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