The answer is peak current = 1.8A
The expression for the Root Mean Square value of current is given as
[tex]$I_{\text {rms }}=\frac{I_{0}}{\sqrt{2}} \text {. }$[/tex]
where, Io, is the peak current
[tex]\begin{aligned}&\text { Power } \quad P=150 \mathrm{~W} . \\&V_{\text {rms }}=120 \mathrm{~V} .\end{aligned}[/tex]
[tex]\begin{aligned}I_{\text {rous }} &=\frac{P}{V} \\&=\frac{150}{120}\end{aligned}[/tex]
[tex]\begin{aligned}& {I_{r m s} }=1.25 \mathrm{~A} \text { : }\\&I_{r m s}=\frac{I_{0}}{\sqrt{2}} \text {. }\end{aligned}[/tex]
[tex]\text { or } \begin{aligned}I_{0} &=I_{r m s} \times \sqrt{2} . \\&=1.25 \times \sqrt{2} . \\I_{0} &=1.7678 \mathrm{~A} . \\1 . e \quad I_{0} &=1.8 \mathrm{~A}\end{aligned}[/tex]
Hence, the peak current is [tex]$1.7678 \mathrm{~A}$[/tex]
[tex]$\text { or } 1.8 \mathrm{~A}$[/tex]
What is peak current?
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