The diode D used in the circuit below is ideal. The voltage drop $V_{a b}$ across the $4 Ω$ resistor is _________Volt

4

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Solution

The correct option is A 4

Let the diode is On.
The diode is replaced by short circuit.
KCL at node A:

$\frac{V_{A} - 3}{2 k Ω} + \frac{V_{A}}{4 k Ω} = 1 m A$
$V_{A} [\frac{1}{2 k Ω} + \frac{1}{4 k Ω}] = \frac{3}{2 k Ω} + 1 m A$
$V_{A} [\frac{6}{8}] = 2.5 m A$
$V_{A} = \frac{8}{6} \times 2.5 m A = 3.33 V$
$I_{D} = \frac{3 - 3.33}{2 k Ω} = - 0.165 m A (- v e i s n o t p o s s i b l e)$
Diode is OFF.
$∴ V_{a b} = 4 \times 1 = 4 v o l t$

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The diode D used in the circuit below is ideal. The voltage drop Vab across the 4 Ω resistor is _________Volt 4

The correct option is A 4 Let the diode is On. The diode is replaced by short circuit. KCL at node A: VA−32 kΩ+VA4 kΩ=1 mA VA[12kΩ+14kΩ]=32kΩ+1 mA VA[68]=2.5mA VA=86×2.5 mA=3.33V ID=3−3.332kΩ=−0.165 mA (−ve is not possible) Diode is OFF. ∴ Vab=4×1=4 volt

The diode D used in the circuit below is ideal. The voltage drop $V_{a b}$ across the $4 Ω$ resistor is _________Volt

4