Question

Find the voltage across $\text{AB}$ in the given circuit. Assume that diode is ideal.

• A. $15.32\text{V}$
• B. $16.32\text{V}$
• C. $15.54\text{V}$
• D. $12.24\text{V}$

Answer 1

The correct option is B $16.32\text{V}$

As the diode is treated ideal, so the resistance of forward biased diode will be zero $(R_f=0)$. It acts as short circuit.

So, $10\text{k}\Omega$ is in parallel with $15\text{k}\Omega$ and the effective resistance across $\text{AB}$ is

$R_{AB}=\frac{10\times 15}{10+15}=\frac{10\times 15}{25}=6\text{k}\Omega$

Now, $6\text{k}\Omega$ is in series with $5\text{k}\Omega$

$\therefore$ Total resistance,

$R_T=6+5=11\text{k}\Omega ,$

Given that, $V=30\text{V}$,

Current drawn from the battery is

$I=\frac{V}{R_T}=\frac{30V}{11\text{k}\Omega }=2.72\text{mA}$

$V_{AB}=I{R}_{AB}=2.72\text{mA}\times 6\text{k}\Omega$

$\therefore V_{AB}=16.32\text{V}$

Hence, option $\left(\text{B}\right)$ is correct.