Consider ideal diodes $D_1$ and $D_2$ with cut-in voltage $V_{\gamma }=0V$ and $v_i\left(t\right)=6\sin \left(\omega t\right)$ in Volt. The maximum voltage (Volt) of the output $v_o\left(t\right)$ is $\text{\_\_\_\_}$ (rounded off to two decimal places).

The correct answer is 4.00.

Analysis of the Circuit Diagram:
Based on the provided circuit diagram, we have:
- An input AC source:
$v_i\left(t\right)=6\sin \left(\omega t\right)$
which has a peak positive voltage of 6 V and a peak negative voltage of -6 V.
- A series input resistor of value:
$R_1=10\text{ k}\Omega$
- A first branch connected in parallel to the output containing an ideal diode D1 in series with a resistor:
$R_2=10\text{ k}\Omega$
and a DC battery of 2 V (with the positive terminal connected to the cathode of D1).
- A second branch in parallel containing an ideal diode D2 in series with a 4 V DC battery (with the positive terminal connected to the cathode of D2).

Step-by-Step State Analysis:

Case 1: When the output voltage is less than 2 V
If the output voltage
$v_o\left(t\right)<2\text{ V}$
both diodes D1 and D2 are reverse-biased (OFF) because the potential at their anodes (the output node) is less than the potentials at their respective cathodes (2 V and 4 V).
Under this condition, no current flows through the parallel branches. Therefore, the output voltage simply tracks the input voltage:
$v_o\left(t\right)=v_i\left(t\right)$
This state continues as the input voltage rises up to 2 V.

Case 2: When the input voltage exceeds 2 V
When the input voltage
$v_i\left(t\right)>2\text{ V}$
the output node voltage attempts to go above 2 V, which forward-biases the ideal diode D1, turning it ON (acting as a short circuit). Meanwhile, as long as the output voltage remains below 4 V, diode D2 remains OFF.
With D1 ON and D2 OFF, the circuit acts as a voltage divider between the input voltage source and the 2 V battery. The output voltage can be calculated as:
$v_o\left(t\right)=2+\left(v_i\right(t)-2)·\frac{R_2}{R_1+R_2}$
Substituting the resistor values
$R_1=10\text{ k}\Omega$
and
$R_2=10\text{ k}\Omega$
gives:
$v_o\left(t\right)=2+\left(v_i\right(t)-2)·\frac{10}{10+10}=2+\frac{v_i\left(t\right)-2}{2}=1+\frac{v_i\left(t\right)}{2}$

Determining the Maximum Output Voltage:
The maximum value of the input signal is
$v_{i,\text{max}}=6\text{ V}$
Substituting this peak value into our voltage divider equation:
$v_{o,\text{max}}=1+\frac{6}{2}=1+3=4\text{ V}$
Since the maximum output voltage is exactly 4 V, diode D2 (which has its cathode at 4 V) is on the verge of conduction but does not turn on to clamp the voltage any further. Thus, the maximum output voltage is 4.00 V.