Answer:
[tex]V=17.9\ Volt[/tex]
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.
[tex]P=I^2R[/tex]
The relation between the voltage and the current is given by Ohm's law:
[tex]V=RI[/tex]
Solving for I and replacing int the first equation
[tex]\displaystyle P=\frac{V^2}{R}[/tex]
Solving for V
[tex]V=\sqrt{PR}[/tex]
[tex]V=\sqrt{0.8\cdot 400}=17.9[/tex]
[tex]\boxed{V=17.9\ Volt}[/tex]
Maximum voltage for given power rating, applied across this resistor without damaging it is 17.9 volts.
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,
[tex]V\propto I[/tex]
Here, (V) is the potential difference and (I) is the current.
It can be written as,
[tex]V=IR[/tex]
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,
[tex]P=I^2\times R\\I=\sqrt{\dfrac{P}{R}}[/tex]
Put this value of current in ohm's law as,
[tex]V=\sqrt{\dfrac{P}{R}}\times R\\V=\sqrt{PR}[/tex]
Put the value of power and current in the above formula,
[tex]V=\sqrt{0.800\times 400}\\V=17.9\rm Volts[/tex]
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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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