The Correct Answer is Option (C)

As per diagram,

Diode D₁ & D₂ are in forward bias i.e. R = 30$\mathrm{\Omega }$ whereas diode D₃ is in reverse bias i.e. R = infinite $\Rightarrow$ Equivalent circuit will be

Applying KVL starting from point A


$-\left(\frac{I_1}{2}\right)\times 30-\left(\frac{I_1}{2}\right)\times 130-I_1\times 20+200=0$ $\Rightarrow$$-$30 I₁ + 200 = 0

I₁ = 2

Option (c)

The Correct Answer is Option (D)

When V_i > 3 volt, V_R > 0

Because diode will be in forward biased state

When V_i $\le$ 3 volt ; V_R = 0

Because diode will be in reverse biased state.

The Correct Answer is Option (C)

Since silicon diode is used so 0.7 Volt is drop across it, only D₁ will conduct.

So current through cell,

I = $\frac{5-0.7}{10}$ = 0.43 A

The Correct Answer is Option (D)

Till input voltage reaches 4V, both diode will stay deactvated so output voltage will be same as input voltage.

When input voltage reach 4V, then 4V diode will get activated and it will transfer only 4V to the output but as a rasistance is present in series with the 4V diode so some voltage drop will happen across the resistor. So output
voltage(V_o) will be = 4V + voltage drop.

When input voltage reach 6V, then 6V diode will get activated and it will transfer only 6V to the output no matter how much input voltage become.

The Correct Answer is Option (D)

1.5 – 0.5 – R $\times$ 10^–2 = 0

$\Rightarrow$ R = 100 $\mathrm{\Omega }$