Question

Figure shows a circuit in which three identical diodes are used. Each diode has forward resistance of $20 \,\Omega$ and infinite backward resistance. Resistors $R_{1}=R_{2}=R_{3}=50\,\Omega$. Battery voltage is $6\, V$. The current through $R_3$ is:

• A. $50\,mA$
• B. $100\,mA$
• C. $60\,mA$
• D. $25\,mA$

Answer 1

Answer: (A)

$50\,mA$

Answer 2

Here, diodes $D_1$ and $D_2$ are forward biased and $D_3$ is reverse biased.
Therefore current through $R_3$
$i=\frac{V}{R '}=\frac{6}{120}=\frac{1}{20}A=50\,mA$