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In the circuit shown for the given input voltage $V_i$, the maximum output voltage $V_0$ is
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Verified Answer
The correct answer is:
$5 \mathrm{~V}$
Given, input voltage,
We know that a $p n$-junction diode is operated in forward bias only.
$\therefore$ For positive half cycle, the circuit will look like,
Here, only $D_2$ is ON and $D_1$ of OFF.
$\therefore$ Voltage drop across $A B=C D=10 \mathrm{~V}$
By using voltage division rule,
Output voltage, across $C E$ (i.e. across $2 \mathrm{k} \Omega$ )
$V_0=\frac{V_i \times 2}{2+2}$
$=\frac{10 \times 2}{4}=5 \mathrm{~V}$
We know that a $p n$-junction diode is operated in forward bias only.
$\therefore$ For positive half cycle, the circuit will look like,
Here, only $D_2$ is ON and $D_1$ of OFF.
$\therefore$ Voltage drop across $A B=C D=10 \mathrm{~V}$
By using voltage division rule,
Output voltage, across $C E$ (i.e. across $2 \mathrm{k} \Omega$ )
$V_0=\frac{V_i \times 2}{2+2}$
$=\frac{10 \times 2}{4}=5 \mathrm{~V}$
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