A temperature sensor has a static transfer fucntion of 0-15 mV- C and time constant of 3-3 sec- If a step change of 22-C to 50-C is applied at t - 0sec- The output voltage in mV at 9sec is

Output voltage at 22^∘C = 0.15 mV^∘C× 22^∘ C = 3.3 mV Output voltage at 50^∘C = 0.15 mV^∘C× 50^∘ C = 7.5 mV V(t) = V_final(t)+ (V_initial (t) V_final(t))e^ t ...

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Byju's Answer

Standard XII

Physics

Multiplication & Division of Errors

A temperature...

Question

A temperature sensor has a static transfer fucntion of $0.15 (\frac{m V}{^{\circ} C})$ and time constant of 3.3 sec. If a step change of $22^{\circ} C$ to $50^{\circ} C$ is applied at t = 0sec. The output voltage (in mV) at 9sec is

6.2

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Solution

The correct option is D 7.2
$Output voltage at 22^{\circ} C = 0.15 \frac{m V}{^{\circ} C} \times 22^{\circ} C = 3.3 m V$

$Output voltage at 50^{\circ} C = 0.15 \frac{m V}{^{\circ} C} \times 50^{\circ} C = 7.5 m V$

$V (t) = V_{f i n a l} (t) + (V_{i n i t i a l} (t) - V_{f i n a l} (t)) e^{- t / τ}$

$V (t) = 7.5 m V + (3.3 m V - 7.5 m V) e^{- t / 3.3}$

$V (t = 9 s e c) = 7.2 (m V)$

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A temperature sensor has a static transfer fucntion of 0.15(mV∘C) and time constant of 3.3 sec. If a step change of 22∘C to 50∘C is applied at t = 0sec. The output voltage (in mV) at 9sec is

The correct option is D 7.2Output voltage at 22∘C=0.15mV∘C×22∘C=3.3mV Output voltage at 50∘C=0.15mV∘C×50∘C=7.5mV V(t)=Vfinal(t)+(Vinitial(t)−Vfinal(t))e−t/τ V(t)=7.5mV+(3.3mV−7.5mV)e−t/3.3 V(t=9sec)=7.2(mV)

A temperature sensor has a static transfer fucntion of $0.15 (\frac{m V}{^{\circ} C})$ and time constant of 3.3 sec. If a step change of $22^{\circ} C$ to $50^{\circ} C$ is applied at t = 0sec. The output voltage (in mV) at 9sec is