Question

The power rating of a 400-Ω resistor is 0.800 W.(a) What is the maximum voltage that can be applied across this resistor without damaging it? Use three significant figures in your answer.

Answer 1

Answer:

$V=17.9\ Volt$

Explanation:

Joule's Law in Electricity

The Joule's law allows us to calculate the power dissipated in a resistor of resistance R through which goes a current I.

$P=I^2R$

The relation between the voltage and the current is given by Ohm's law:

$V=RI$

Solving for I and replacing int the first equation

$\displaystyle P=\frac{V^2}{R}$

Solving for V

$V=\sqrt{PR}$

$V=\sqrt{0.8\cdot 400}=17.9$

$\boxed{V=17.9\ Volt}$

Answer 2

Maximum voltage for given power rating, applied across this resistor without damaging it is 17.9 volts.

What is the Ohm's law?

Ohm's law states that for a flowing current the potential difference of the circuit is directly proportional to the current flowing in it. Thus,

$V\propto I$

Here, (V) is the potential difference and (I) is the current.

It can be written as,

$V=IR$

Here, (R) is the resistance of the circuit.

Given information-

The value of resistance is 400-Ω.

The value of power rating is 0.800 W.

By the Joule's law, the power of a circuit is equal to the product of the  square of the current flowing in it and the resistance. It can be given as,

$P=I^2\times R\\I=\sqrt{\dfrac{P}{R}}$

Put this value of current in ohm's law as,

$V=\sqrt{\dfrac{P}{R}}\times R\\V=\sqrt{PR}$

Put the value of power and current in the above formula,

$V=\sqrt{0.800\times 400}\\V=17.9\rm Volts$

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Thus the maximum voltage that can be applied across this resistor without damaging it is 17.9 volts.

Learn more about the Ohm's law here;

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