The voltage $\mathrm{V}_0$ in the network shown is

• Correct answer
  A

  $\mathrm{V}_0=11.3 \mathrm{~V}$
• B

  $\mathrm{V}_0=9.8 \mathrm{~V}$
• C

  $\mathrm{V}_0=12.0 \mathrm{~V}$
• D

  $\mathrm{V}_0=0.7 \mathrm{~V}$

Solution

In the given circuit, Potential barrier of LED > Potential barrier of diode So, the equivalent circuit diagram is $\begin{aligned} & \therefore \text { By KVL, } \\ & 12-0.7=V_0 \\ & \therefore V_0=11.3 \mathrm{~V} \end{aligned}$

Hence, the correct answer is option: 'A'.