Consider the circuit shown below,
If the emitter current of the transistor is $I_{E} = 2 m A$ , then the value of $R_{1} =_____k Ω$

53.6

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Solution

The correct option is A 53.6
$I_{B} = \frac{I_{E}}{1 + β} = \frac{2 m A}{1 + 49} = 40 μ A$

$I_{C} = β . I_{B} = 49 \times 40 μ A = 1.96 m A$

$V_{B} = V_{B E} + I_{E} R_{E} = 0.7 + (2 \times 10^{- 3} \times 100) = 0.9 V$

$I_{R 2} = \frac{V_{B}}{R_{2}} = \frac{0.9}{20 k}$

$I_{R 2} = 45 μ A$

$I_{R 1} = I_{B} + I_{R 2} = 40 + 45 = 85 m u A$

$V_{A} = 12 - (I_{R 1} + I_{C}) R_{C}$

$= 12 - (85 μ + 1.96 m A) \times 3.2 \times 10^{3} = 5.456 V o l t$

$V_{B} = 0.9$

$R_{1} = \frac{5.456 - 0.9}{85 μ A} = 53.60 k Ω$

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Consider the circuit shown below, If the emitter current of the transistor is IE=2mA , then the value of R1=_____kΩ 53.6

The correct option is A 53.6IB=IE1+β=2 mA1+49=40 μA IC=β.IB=49×40μA=1.96 mA VB=VBE+IERE=0.7+(2×10−3×100)=0.9V IR2=VBR2=0.920k IR2=45 μA IR1=IB+IR2=40+45=85 muA VA=12−(IR1+IC)RC =12−(85 μ+1.96 mA)×3.2×103=5.456 Volt VB=0.9 R1=5.456−0.985 μA=53.60 kΩ

Consider the circuit shown below,
If the emitter current of the transistor is $I_{E} = 2 m A$ , then the value of $R_{1} =_____k Ω$

53.6