The voltage $\mathrm{V}_0$ in the network shown is
The voltage $\mathrm{V}_0$ in the network shown is
$\mathrm{V}_0=11.3 \mathrm{~V}$
$\mathrm{V}_0=9.8 \mathrm{~V}$
$\mathrm{V}_0=12.0 \mathrm{~V}$
$\mathrm{V}_0=0.7 \mathrm{~V}$
Solution
In the given circuit,
Potential barrier of LED > Potential barrier of diode
So, the equivalent circuit diagram is
$\begin{aligned}
& \therefore \text { By KVL, } \\
& 12-0.7=V_0 \\
& \therefore V_0=11.3 \mathrm{~V}
\end{aligned}$
