Question
Prove that the power dissipated in an ideal resistor connected to an ac source is $\frac{\text{V}^2_{\text{eff}}}{\text{R}}.$
✓
Answer
Power in ac circuit, P $=\text{V}_{\text{rms}}\text{i}_{\text{rms}}\cos\varphi$
As rms values of current and voltage are also caled effective values i.e.
$\text{P = V}_{\text{eff}}\text{I}_{\text{eff}}\cos\varphi \ ...(\text{i})$
But $\cos\varphi=\text{power factor}=\frac{\text{R}}{\text{Z}}$
In a purely resistive circuit Z $=\text{R},\cos\varphi=1$
and $\text{i}_{\text{eff}}=\frac{\text{V}_{\text{eff}}}{\text{Z}}=\frac{\text{V}_{\text{eff}}}{\text{R}}$
Substituting these value in (i), we get
$\text{P}=\text{V}_{\text{eff}}.\frac{\text{V}_{\text{eff}}}{\text{R}}\times1=\frac{\text{V}^2_{\text{eff}}}{\text{R}}$
As rms values of current and voltage are also caled effective values i.e.
$\text{P = V}_{\text{eff}}\text{I}_{\text{eff}}\cos\varphi \ ...(\text{i})$
But $\cos\varphi=\text{power factor}=\frac{\text{R}}{\text{Z}}$
In a purely resistive circuit Z $=\text{R},\cos\varphi=1$
and $\text{i}_{\text{eff}}=\frac{\text{V}_{\text{eff}}}{\text{Z}}=\frac{\text{V}_{\text{eff}}}{\text{R}}$
Substituting these value in (i), we get
$\text{P}=\text{V}_{\text{eff}}.\frac{\text{V}_{\text{eff}}}{\text{R}}\times1=\frac{\text{V}^2_{\text{eff}}}{\text{R}}$
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