Step 1: Understand the circuit.
The given circuit is an op-amp with diodes and a Zener diode arrangement, forming a limiter.
- For positive half cycles, \(D_1\) conducts and \(D_Z\) (Zener) enters breakdown after 5 V.
- For negative half cycles, \(D_2\) conducts.
Step 2: Positive cycle.
When input \(V_{in}\) is positive and large:
- The op-amp drives its output positive.
- Once the Zener diode conducts in reverse at \(5 \, V\), additional drop across the 1k\(\Omega\) resistor allows the output to rise up to approximately \(+10 \, V\).
Thus, the positive maximum output is \(\; +10 \, V\).
Step 3: Negative cycle.
When input \(V_{in}\) is negative and large:
- The diode \(D_2\) clamps the negative excursion.
- This clamps the voltage to approximately \(-5 \, V\).
Thus, the minimum output is \(\; -5 \, V\).
Step 4: Final range.
Therefore,
\[
V_O \in [-5, +10]
\]
Final Answer:
\[
\boxed{+10 \, V \; \text{and} \; -5 \, V}
\]
A continuous time periodic signal \( x(t) \) is given by: \[ x(t) = 1 + 2\cos(2\pi t) + 2\cos(4\pi t) + 2\cos(6\pi t) \] If \( T \) is the period of \( x(t) \), then evaluate: \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt \quad {(round off to the nearest integer).} \]
The maximum percentage error in the equivalent resistance of two parallel connected resistors of 100 \( \Omega \) and 900 \( \Omega \), with each having a maximum 5% error, is: \[ {(round off to nearest integer value).} \]
Consider a distribution feeder, with \( R/X \) ratio of 5. At the receiving end, a 350 kVA load is connected. The maximum voltage drop will occur from the sending end to the receiving end, when the power factor of the load is: \[ {(round off to three decimal places).} \]
In the circuit with ideal devices, the power MOSFET is operated with a duty cycle of 0.4 in a switching cycle with \( I = 10 \, {A} \) and \( V = 15 \, {V} \). The power delivered by the current source, in W, is: \[ {(round off to the nearest integer).} \] 
The induced emf in a 3.3 kV, 4-pole, 3-phase star-connected synchronous motor is considered to be equal and in phase with the terminal voltage under no-load condition. On application of a mechanical load, the induced emf phasor is deflected by an angle of \( 2^\circ \) mechanical with respect to the terminal voltage phasor. If the synchronous reactance is \( 2 \, \Omega \), and stator resistance is negligible, then the motor armature current magnitude, in amperes, during loaded condition is closest to: \[ {(round off to two decimal places).} \]