The Correct Answer is Option (D)

Case 1 :

Ge need 0.3 V for start conducting and Si need 0.7 V for start conducting.

So, conduction will be start by Ge.

Out of 12 V, 0.3V will be consumed by Ge.

$\therefore$  V₀₁ = 12 $-$ 0.3 = 11.7 V

Case 2 :

Here as Ge diode is in reverse bias so no conduction will happen through Ge diode

As Si diode is in forward bias so conduction will happen through Si diode.

Here Si diode consume 0.7 V.

So, V₀₂ = 12 $-$ 0.7 = 11.3 V

$\therefore$  Value of V₀ changes by

= 11.7 $-$ 11.3 = 0.4 V

The Correct Answer is Option (D)

Note

From the circuit you can see the diode is in forward bias, and to start current flow through the circuit. Accross silicon diode the voltage drop should be 0.7 V and for Germenium diode the voltage drop is 0.3 V. Then diode will start working and current will flow through the circuit.

So, the net voltage available accros the registor is (V_net) = 3 $-$ 0.7 = 2.3 V

$\therefore$ Current flow through circuit = $\frac{2.3}{200}$ = 11.5 mA

The Correct Answer is Option (B)

Given : Resistance $R=100$ kilo $ohm=100$ $\times {10}^3\mathrm{\Omega }$

$\therefore$ Capacitance $C=250$ picofarad $=250\times {10}^{-12}F$

$\tau =RC=100\times {10}^3\times 250\times {10}^{-12}\sec$

$=2.5\times {10}^7\times {10}^{-12}\sec$

$=2.5\times {10}^{-5}\sec$

The higher frequency which can be detected with tolerable distortion is

$f=\frac{1}{2\pi m_{a}RC}$

$=\frac{1}{2\pi \times 0.6\times 2.5\times {10}^{-5}}Hz$

$=\frac{100\times {10}^4}{2.5\times 1.2\pi }Hz$

$=10.61KHz$

This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less than the rate of decay modulated signal voltage for proper detection of modulated signal.

The Correct Answer is Option (A)

The current will flow through $R_L$ when the diode is forward biased.

The Correct Answer is Option (B )

$D_2$ is forward biased whereas $D_1$ is reversed biased.

So effective resistance of the circuit

$R=4+2=6\mathrm{\Omega }$

$\therefore$ $i=\frac{12}{6}=2A$