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(A) Given the comparator circuit,the output $V_{o}$ is determined by the inputs $V_{+}$ and $V_{-}$.$V_{o} = \begin{cases} +10 \, V & 	ext{if } V_{+}

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(A) Given the comparator circuit,the output $V_{o}$ is determined by the inputs $V_{+}$ and $V_{-}$.$V_{o} = \begin{cases} +10 \, V & 	ext{if } V_{+} > V_{-} \ -10 \, V & 	ext{if } V_{+} From the circuit,the non-inverting input $V_{+}$ is connected to a voltage divider formed by two resistors $R$ in series connected to $V_{o}$,so $V_{+} = V_{o} \left( \frac{R}{R+R} ight) = \frac{V_{o}}{2}$.The inverting input $V_{-}$ is connected to the capacitor $C$,so $V_{-} = V_{C}$.When $V_{o} = +10 \, V$,then $V_{+} = +5 \, V$. The capacitor charges towards $+10 \, V$ through the resistor $R$. As soon as $V_{C}$ exceeds $+5 \, V$,$V_{-} > V_{+}$,causing the output to switch to $-10 \, V$.When $V_{o} = -10 \, V$,then $V_{+} = -5 \, V$. The capacitor discharges towards $-10 \, V$. As soon as $V_{C}$ drops below $-5 \, V$,$V_{-} This continuous switching results in a square wave output oscillating between $+10 \, V$ and $-10 \, V$.

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