Figure shows a circuit in which three identical diodes are used. Each diode has forward resistance of $20…
Figure shows a circuit in which three identical diodes are used. Each diode has forward resistance of $20 \Omega$ and infinite backward resistance. Resistors $\mathrm{R}_1=\mathrm{R}_2=\mathrm{R}_3=50 \Omega$. Battery voltage is $6 \mathrm{~V}$. The current through $\mathrm{R}_3$ is :
$50 \mathrm{~mA}$
$100 \mathrm{~mA}$
$60 \mathrm{~mA}$
$25 \mathrm{~mA}$
Solution
Here, diodes $D_1$ and $D_2$ are forward biased and $\mathrm{D}_3$ is reverse biased. Therefore current through $\mathrm{R}_3$
$
\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}=\frac{6}{120}=\frac{1}{20} \mathrm{~A}=50 \mathrm{~mA}
$
