Question

Considering the circuit and data given in the diagram, calculate the currents flowing in the diodes $D_1$ and $D_2$. Forward resistance of $D_1$ and $D_2$ is $20\Omega$.

• A. $\frac{1}{220}\text{A},\frac{1}{220}\text{A}$
• B. $\frac{1}{220}\text{A},\frac{1}{110}\text{A}$
• C. $\frac{1}{110}\text{A},\frac{1}{220}\text{A}$
• D. $\frac{1}{110}\text{A},\frac{1}{110}\text{A}$

Answer 1

The correct option is A $\frac{1}{220}\text{A},\frac{1}{220}\text{A}$

Since the positive terminal of the battery is connected to $p-$ type of both diodes $D_1$ and $D_2$, those are forward biased.

These diodes are replaced by with their forward resistance as shown in figure,


The effective resistance of $20\Omega$ and $20\Omega$ in parallel,

$\frac{1}{R}=\frac{1}{20}+\frac{1}{20}$

$\Rightarrow R=\frac{20}{2}=10\Omega$

Therefore, total current $I$ in the circuit

$I=\frac{V}{R}=\frac{1}{100+10}=\frac{1}{110}\text{A}$

$\Rightarrow I_1=I_2=\frac{1}{2}\times \frac{1}{110}=\frac{1}{220}\text{A}$

Hence, option $\left(A\right)$ is correct.