An electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source?
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$\text{E}=12\text{volt}$
$\text{i}^2\text{Rt}=\text{i}^2_\text{rms}\text{RT}$
$\Rightarrow\ \frac{\text{E}^2}{\text{R}^2}=\frac{\text{E}^2_\text{rms}}{\text{R}^2}$
$\Rightarrow\ \text{E}^2=\frac{\text{E}_0^2}{2}$
$\Rightarrow\ \text{E}_0{^2}=2\text{E}^2$
$\Rightarrow\ \text{E}_0{^2}=2\times12^2=2\times144$
$\Rightarrow \text{E}_0=\sqrt{2\times144}=16.97\approx17\text{V}$
$\text{i}^2\text{Rt}=\text{i}^2_\text{rms}\text{RT}$
$\Rightarrow\ \frac{\text{E}^2}{\text{R}^2}=\frac{\text{E}^2_\text{rms}}{\text{R}^2}$
$\Rightarrow\ \text{E}^2=\frac{\text{E}_0^2}{2}$
$\Rightarrow\ \text{E}_0{^2}=2\text{E}^2$
$\Rightarrow\ \text{E}_0{^2}=2\times12^2=2\times144$
$\Rightarrow \text{E}_0=\sqrt{2\times144}=16.97\approx17\text{V}$
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