Question

If $V_1>V_2$ , $r$ is the resistance offered by the diode in forward bias then current through the diode is

• A. $0$
• B. $\frac{V_1+V_2}{R+r}$
• C. $\frac{V_1-V_2}{R+r}$
• D. $\frac{V_1+V_2}{R}$

Answer 1

The correct option is B $\frac{V_1+V_2}{R+r}$


The potential of point $\text{A}$ is $V_1$ while the potential at point $\text{B}$ is $-V_2$.

Therefore, the potential difference through $\text{AB}$,

$V_{AB}=V_1-(-V_2)$

$\Rightarrow V_{AB}=V_1+V_2$

Since the potential at $\text{A}$ is greater than that of $\text{B}$, the diode is forward biased.

Given that, the forward resistance of the diode is $r$ and the diode is in series with $R$,

$R_{AB}=R+r$

According to ohm's law

$V_{AB}=i{R}_{AB}$

$(V_1+V_2)=i(R+r)$

$\Rightarrow i=\left(\frac{V_1+V_2}{R+r}\right)$

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