Search any question & find its solution
Question:
Answered & Verified by Expert
In the circuit shown in figure given below, if the diode forward voltage drop is $0.3 \mathrm{~V}$, the voltage difference between $\mathrm{A}$ and $\mathrm{B}$ is
Options:
Solution:
2576 Upvotes
Verified Answer
The correct answer is:
$2.3 \mathrm{~V}$
$2.3 \mathrm{~V}$
Let the potential difference between $\mathrm{A}$ and $\mathrm{B}$ is $\mathrm{V}$, Given here $r_1=5 \mathrm{k} \Omega$ and $r_2=5 \mathrm{k} \Omega$ are resistance in series connection.
So,
$$
\begin{aligned}
&\left.\mathrm{V}_{A B}-0.3=\left[\left(\mathrm{r}_1+\mathrm{r}_2\right) 10^3\right] \times\left(0.2 \times 10^{-3}\right)\right] \\
&\quad[\because \mathrm{V}=\mathrm{ir}] \\
&\left(\mathrm{V}_{A B}-0.3\right)=10 \times 10^3 \times 0.2 \times 10^{-3}=2 \\
&\text { So, } \mathrm{V}_{\mathrm{AB}}=2+0.3=2.3 \mathrm{~V}
\end{aligned}
$$
So,
$$
\begin{aligned}
&\left.\mathrm{V}_{A B}-0.3=\left[\left(\mathrm{r}_1+\mathrm{r}_2\right) 10^3\right] \times\left(0.2 \times 10^{-3}\right)\right] \\
&\quad[\because \mathrm{V}=\mathrm{ir}] \\
&\left(\mathrm{V}_{A B}-0.3\right)=10 \times 10^3 \times 0.2 \times 10^{-3}=2 \\
&\text { So, } \mathrm{V}_{\mathrm{AB}}=2+0.3=2.3 \mathrm{~V}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.