Consider the circuit shown in the figure below-If diode D1 and D2 are made up of same material with the cut-in voltage VY-0-7 V- then the value of current I is equal to

When D_2 is ON then the value of nbsp; nbsp;V_0 will be V_0=3 0.7 V=2.3 V Hence, D_1 will be OFF. Thus, nbsp; The current, I=2.3 ( 3)5×10^ 3=5.35×1 ...

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Standard X

Physics

Diodes

Consider the ...

Question

Consider the circuit shown in the figure below.

$\text{If diode }D_1 \text{ and }D_2 \text{ are made up of same material with the cut-in voltage }V_Y=0.7\text{ V, then the value of current I is equal to}$

0.59 mA

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0.99 mA

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0.46 mA

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1.06 mA

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Solution

The correct option is D 1.06 mA
$When D_{2} is ON then the value of V_{0} will be V_{0} = 3 - 0.7 V = 2.3 V$
$Hence, D_{1} will be OFF.$
$Thus, The current, I = \frac{2.3 - (- 3)}{5} \times 10^{- 3} = \frac{5.3}{5} \times 10^{- 3} = 1.06 m A$

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Consider the circuit shown in the figure below. \(\text{If diode }D_1 \text{ and }D_2 \text{ are made up of same material with the cut-in voltage }V_Y=0.7\text{ V, then the value of current I is equal to}\)

The correct option is D 1.06 mAWhen D2 is ON then the value of V0 will be V0=3−0.7V=2.3V Hence, D1 will be OFF. Thus, The current, I=2.3−(−3)5×10−3=5.35×10−3=1.06mA

Consider the circuit shown in the figure below.

\(\text{If diode }D_1 \text{ and }D_2 \text{ are made up of same material with the cut-in voltage }V_Y=0.7\text{ V, then the value of current I is equal to}\)

The correct option is D 1.06 mA
$When D_{2} is ON then the value of V_{0} will be V_{0} = 3 - 0.7 V = 2.3 V$
$Hence, D_{1} will be OFF.$
$Thus, The current, I = \frac{2.3 - (- 3)}{5} \times 10^{- 3} = \frac{5.3}{5} \times 10^{- 3} = 1.06 m A$