In the circuit shown below-given that -100-VBE-0-7 V- The value of drain voltage Vb is

According to voltage division rule, V_B=15× R_2R_1+R_2=15× 2.5k10k=3.75 V Applying KVL in base emitter loop of transistor; V_B+V_BE+V_E=0 V_E=V_B V_BE=3.75 ...

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Standard XII

Physics

Full Wave Rectifier

In the circui...

Question

In the circuit shown below,
given that $β = 100, V_{B E} = 0.7$ V. The value of drain voltage $V_{b}$ is

9.4 V

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7.5 V

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8.9 V

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10.2 V

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Solution

The correct option is C 8.9 V
According to voltage division rule,

$V_{B} = \frac{15 \times R_{2}}{R_{1} + R_{2}} = \frac{15 \times 2.5 k}{10 k} = 3.75 V$

Applying KVL in base emitter loop of transistor;

$- V_{B} + V_{B E} + V_{E} = 0$
$V_{E} = V_{B} - V_{B E} = 3.75 - 0.7$
= 3.05 V

$I_{E} = \frac{V_{E}}{R_{E}} = \frac{3.05}{1 k} = 3.05$ mA

$I_{E} = I_{C} = I_{D} = I_{S} = 3.05$ mA

$\frac{V_{D D} - V_{D}}{R_{D}} = I_{D}$

$V_{D} = 15 - 6.1 = 8.9 V$

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In the circuit shown below, given that β=100,VBE=0.7 V. The value of drain voltage Vb is

The correct option is C 8.9 VAccording to voltage division rule, VB=15×R2R1+R2=15×2.5k10k=3.75 V Applying KVL in base emitter loop of transistor; −VB+VBE+VE=0 VE=VB−VBE=3.75−0.7 = 3.05 V IE=VERE=3.051k=3.05 mA IE=IC=ID=IS=3.05 mA VDD−VDRD=ID VD=15−6.1=8.9 V

In the circuit shown below,
given that $β = 100, V_{B E} = 0.7$ V. The value of drain voltage $V_{b}$ is