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RC transients-3

In the microp...

Question

In the microprocessor controlled measurement scheme shown in the figure, $R_{x}$ is the unknown resistance to be measured, while $R_{r e f}$ and $C_{r e f}$ are known. $C_{r e f}$ is charged from voltage $V_{L}$ to $V_{H}$ (by a constant DC voltage source $V_{S}$ ), once through $R_{r e f}$ in $T_{r e f}$ seconds and then discharged to $V_{L}$ . It is again charged from voltage $V_{L}$ to $V_{H}$ through $R_{x}$ in $T_{x}$ seconds.

if $T_{x}$ = $k T_{r e f}$ then

R_{x} = k R_{r e f} (1 - \frac{V_{L}}{V_{H}})

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R_{x} = k R_{r e f} l n (\frac{V_{H}}{V_{L}})

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R_{x} = k R_{r e f}

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R_{x} = R_{r e f} l n k

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Solution

The correct option is C $R_{x} = k R_{r e f}$
$v_{c} (t) = [v_{c} (0^{+}) - v_{c} (\infty)] e^{- t / R C}$

$v_{H} = (V_{L} - V_{s}) e^{- T_{r e f} / R_{r e f} C_{r e f}}$ ....(i)

$v_{H} = (V_{L} - V_{s}) e^{- T_{x} / R_{x} C_{r e f}}$ ....(ii)

Comparing equations (i) and (ii),

$- \frac{T_{r e f}}{R_{r e f} C_{r e f}}$ = $- \frac{T_{x}}{R_{x} C_{r e f}}$

$f o r T_{x} = k T_{r e f}$

$R_{x} = k R_{r e f}$

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The correct option is C Rx=kRrefvc(t)=[vc(0+)−vc(∞)]e−t/RC vH=(VL−Vs)e−Tref/RrefCref ....(i) vH=(VL−Vs)e−Tx/RxCref ....(ii) Comparing equations (i) and (ii), −TrefRrefCref = −TxRxCref for Tx=kTref Rx=kRref