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If the input $\mathrm{AC}$ voltage is $10 \mathrm{Vrms}$, find the maximum voltage across the diode of a half wave rectifier with capacitor input filter.
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Solution:
2864 Upvotes
Verified Answer
The correct answer is:
$14.1 \mathrm{~V}$
Concept:
Low-voltage rectifiers require a step-down transformer to reduce the strength of AC voltage.
$\frac{V_1}{V_2}=\frac{N_1}{N_2}$
A step-down transformer is necessary for:
$\mathrm{Vm}$ is the peak voltage
Calculation:
Given
$\begin{aligned}
& \mathrm{Vrms}=10 \mathrm{~V} \\
& \mathrm{Vm}=V_{r m s} \times \sqrt{2} \\
& =10 \times \sqrt{2} \\
& =14.1 \mathrm{~V}
\end{aligned}$
Low-voltage rectifiers require a step-down transformer to reduce the strength of AC voltage.
$\frac{V_1}{V_2}=\frac{N_1}{N_2}$
A step-down transformer is necessary for:
$\mathrm{Vm}$ is the peak voltage
Calculation:
Given
$\begin{aligned}
& \mathrm{Vrms}=10 \mathrm{~V} \\
& \mathrm{Vm}=V_{r m s} \times \sqrt{2} \\
& =10 \times \sqrt{2} \\
& =14.1 \mathrm{~V}
\end{aligned}$
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