Answer :
The relation between rms voltage and peak voltage is given as,
[tex]V_{mn} = \frac{V_{0} }{\sqrt{2} }[/tex]
here, [tex]V_{0}[/tex] is the peak voltage.
Rewrite the above equation in term of peak voltage,
Vā = [tex]\sqrt{2V_{mn} }[/tex]
The relation between rms current and peak current is given as,
[tex]I_{mn} =\frac{I_{0} }{\sqrt{2} }[/tex]
Here, [tex]I_{mn}[/tex]is the rms current and [tex]I_{0}[/tex], is the peak current.
Rewrite the above equation in term of peak current,
[tex]I_{0} =\sqrt{2I_{mn} }[/tex]
The relation between average AC power, peak voltage and peak current is given as,
[tex]P_{ave} =\frac{1}{2} I_{0}V_{0}[/tex]
Here, V, is the peak voltage and I, is the peak current.
Substitute ā2Vmn for Vand ā21 for Iin equation, rms rms
[tex]P_{ave} =\frac{1}{2} I_{0}V_{0}[/tex]
solve,
[tex]P_{ave} =\frac{1}{2} (\sqrt{2I_{mn} } )(\sqrt{2V_{mn} })\\ I_{mn}V_{mn}[/tex]
substitute 500W for Pawe and 120V fpr Vrmn in eq.
[tex]P_{ave} = I_{mn} V_{mn} \\\\[/tex]
solve,
[tex]I_{mn} =\frac{500w}{120v} \\[/tex] =4.167A
4.167A =[tex]A=\frac{I_{0} }{\sqrt{2} } \\I_{0}=\sqrt{2(4.167A)} \\ =5.89A\\[/tex]
what is current ?
In a full electrical circuit, current refers to the speed at which electrons move past a certain location. Current Means flow, at its most fundamental level.
The international unit for measuring current is an ampere (AM-pir), sometimes known as an amp. It describes how many electrons (sometimes referred to as "electrical charge) travel across a certain location in a circuit during a specified period of time.
A voltage-producing energy source, such as a battery, is included into the circuit. Without voltage, electrons in a wire travel unevenly and erratically, and current cannot flow. Electrons are propelled in a single direction by pressure produced by voltage.
Any device (a load) attached to the circuit receives energy from the circuit by way of a closed, conductive loop formed by the circuit. When a switch is switched to the ON, or closed, position, a circuit is closed (complete).
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