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Half Wave Rectifier

Consider a no...

Question

Consider a non-ideal voltage source whose output voltage is measured by a non-ideal voltmeter as shown below.

Let $V_{e}$ be the difference between $V_{s}$ and the measured voltage. $V_{e} / V_{s}$ is function of

A

R_{m} o n l y

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B

R_{s} o n l y

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C

R_{s} / R_{m}

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D

R_{m} - R_{s}

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Solution

The correct option is C $R_{s} / R_{m}$
Measured voltage $= V_{0}$

$V_{0} = \frac{R_{m}}{R_{m} + R_{s}} V_{s}$

$V_{e} = V_{s} - \frac{R_{m}}{R_{m} + R_{s}} V_{s}$

$= \frac{V_{s} (R_{m} + R_{s}) - V_{s} (R_{m})}{R_{m} + R_{s}} = \frac{V_{s} R_{s}}{R_{m} + R_{s}}$

$∴ \frac{V_{e}}{V_{s}} = \frac{R_{s}}{R_{m} + R_{s}} = \frac{R_{s}}{R_{s} (\frac{R_{m}}{R_{s}} + 1)}$

$= \frac{1}{\frac{R_{m}}{R_{s}} + 1} = \frac{1}{\frac{1}{R_{s} / R_{m}} + 1}$

$∴ Clearly, \frac{V_{e}}{V_{s}} = f (\frac{R_{s}}{R_{m}})$

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