Figure shows a circuit in which three identical diodes are used. Each diode has forward resistance of 20 Ω and infinite backward resistance. Resistors R1=R2=R3=50 Ω. Battery voltage is 6 V. The current through R3 is:

Question

Figure shows a circuit in which three identical diodes are used. Each diode has forward resistance of 20 Ω and infinite backward resistance. Resistors R1=R2=R3=50 Ω. Battery voltage is 6 V. The current through R3 is: (A) 50 mA (B) 100 mA (C) 60 mA (D) 25 mA

Tardigrade Admin · Accepted Answer

Here, diodes D1 and D2 are forward biased and D3 is reverse biased. Therefore current through R3 i=(V/R ')=(6/120)=(1/20)A=50 mA

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

$50 m A$

$100 m A$

$60 m A$

$25 m A$

Here, diodes $D_{1}$ and $D_{2}$ are forward biased and $D_{3}$ is reverse biased.
Therefore current through $R_{3}$
$i = \frac{V}{R ^{'}} = \frac{6}{120} = \frac{1}{20} A = 50 m A$

Q. Figure shows a circuit in which three identical diodes are used. Each diode has forward resistance of 20Ω and infinite backward resistance. Resistors R1​=R2​=R3​=50Ω. Battery voltage is 6V. The current through R3​ is:

50mA

100mA

60mA

25mA