In the following circuit of PN junction diodes $D_{1}, D_{2} a n d D_{3}$ are ideal $i$ is:-

\frac{E}{R}

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\frac{E}{2 R}

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\frac{2 e}{3 R}

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Zero

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Solution

The correct option is A $\frac{E}{R}$
Diodes $D_{1}$ and $D_{3}$ are forward biased and $D_{2}$ is reverse biased so the circuit can be redrawn as follows.
$\Rightarrow i = \frac{E}{R}$

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Similar questions

Q. In the following circuit of PN junction diodes

D_{1}, D_{2} a n d D_{3}

are ideal then i is

Q. In the following circuits p-n junction diodes

D_{1}, D_{2}

and

D_{3}

are ideal. For the following potential of A and B, the correct increasing order of resistance between A and B will be
i.

- 10 V, - 5 V

ii.

- 5 V, - 10 V

iii.

- 4 V, - 12 V

Two resistors of resistance 2R and 3R are connected in series with a battery, which has an emf of E and an internal resistance R. What is the potential difference across the resistor of resistance 2R?

Q. Which of the following must be zero in an unbiased PN junction diode?

Q. In the circuit shown in figure,

R_{1} = R_{2} = R_{3} = R

Table 1	Table - 2
(a) current through $R_{1}$	(p) E/R
(b) current through $R_{2}$	(q) 2E/R
(c) current through $R_{3}$	(r) E/2R
	(s) Zero

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In the following circuit of PN junction diodes D1,D2andD3 are ideal i is:-

The correct option is A ERDiodes D1 and D3 are forward biased and D2 is reverse biased so the circuit can be redrawn as follows.⇒i=ER

In the following circuit of PN junction diodes $D_{1}, D_{2} a n d D_{3}$ are ideal $i$ is:-

The correct option is A $\frac{E}{R}$
Diodes $D_{1}$ and $D_{3}$ are forward biased and $D_{2}$ is reverse biased so the circuit can be redrawn as follows.
$\Rightarrow i = \frac{E}{R}$