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14. Semicondutor electronics — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics14. Semicondutor electronics1 Mark
MCQ
The box in the circuit below has two inputs marked $V_{+}$and $V_{-}$and a single output marked $V_{o}$. The output obeys $V_{o}=\left\{\begin{array}{l}+10 V \text { if } V_{+} > V_{-} \\ -10 V \text { if } V_{+} < V_{-}\end{array}\right.$
  • ✓

  • B

  • C

  • D

✓

Answer

Correct option: A.

a
$(a)$ Given, $V_{o}=\left\{\begin{array}{l}+10 V \text { if } V_{+} > V_{-} \\ -10 V \text { if } V_{+} < V_{-}\end{array}\right.$

Initially, when the capacitor is fully charged,

$V_{-}=V_{C}=V_{o}$

From the circuit,

$V_{+}=V_{o}\left(\frac{R}{R+R}\right)=\frac{V_{0}}{2}$

As, $V_{-}>V_{+}$, so $\quad V_{o}=-10 \,V$

For $V_{o}=-10 V , \quad V_{+}=\frac{-10}{2}=-5 \,V$

and $\quad V_{-}=-10 \,V$

i.e., $\quad V_{+}>V_{-}$

and $\quad V_{o}=+10 \,V$

As, the values are fixed, so correct graph is shown in option $(a)$

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