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Question

# Consider the circuit shown in the figure below. The switch in the circuit is initially closed for a long time, at t = 0 it is opened. Then which of the following statement is true. (Assume forward drop of a zener diode = 0.7 V).

A
It makes a transition from +5.7 V to 5.7 at t=13 μsec
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B
It makes a transition from +5.7Vto5.7Vatt=14.23 μsec
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C
It makes a transition from +5.7 V at t=3 μsec
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D
It makes a transistion from +5.7 V to 5.7 V at t=1.29 μsec.
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Solution

## The correct option is B It makes a transition from +5.7Vto−5.7Vatt=14.23 μsecIn the circuit, the capacitor starts charginng from 0 V (as the switch was initially closed) towards the steady state alue of 0 V. Now, when he switch is flipped open, the capacitor will charge upto 20 V. ∴Vc(t)=Vc(∞)−[Vc(0)−Vc(∞)]e−t/RC RC=1×103×0.01×10−6=10 μ sec Vc(t)=20(1−e−t/RC) Voltage at non-inverting amplifier is obtained as V+=V0×100 kΩ(10+100) kΩ V+=V0×100110=V0×1011 ∵ Initially V− was equal to -10 V, thus V0=+5.7 V. Thus, now capacitor will start charging as soon as the switch is opened. Thus, V−=VC−10 V or, VC=V−+10 V now, V−=V+=5.7 V×1011[∵ the op-amp will switch] thus, VC=10×5.711+10 now, VC=20(1−e−t/RC) ∴20(1−e−t/RC)=10+5711 1−e−t/RC=12+57220 1−e−t/RC=0.7590 1−e−t/RC=0.2409 T=14.23 μsec Hence, the output voltage wave will be,

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