For the transistor circuit shown below,if bet | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Applying Kirchhoff's Voltage Law $(KVL)$ to the input loop:$V_{BB} - I_B R_B - V_{BE} = 0$$5\ V - I_B (8.6\ k\Omega) - 0.7\ V = 0$$I_B = \frac{4.3

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) Applying Kirchhoff's Voltage Law $(KVL)$ to the input loop:$V_{BB} - I_B R_B - V_{BE} = 0$$5\ V - I_B (8.6\ k\Omega) - 0.7\ V = 0$$I_B = \frac{4.3\ V}{8.6 	imes 10^3\ \Omega} = 0.5 	imes 10^{-3}\ A = 0.5\ mA$Now,calculate the collector current $I_C$:$I_C = \beta I_B = 100 	imes 0.5\ mA = 50\ mA = 0.05\ A$Applying $KVL$ to the output loop:$V_{CC} - I_C R_L - V_{CE} = 0$$18\ V - (0.05\ A 	imes 100\ \Omega) - V_{CE} = 0$$18\ V - 5\ V - V_{CE} = 0$$V_{CE} = 13\ V$

For the transistor circuit shown below,if $\beta = 100$ and the voltage drop between the emitter and base is $0.7\ V$,then the value of $V_{CE}$ will be.....$V$.

Similar Questions

Vegetative propagation in Pistia occurs by:

Which of the products is formed when acetone is reacted with barium hydroxide solution?

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

Which is the smallest blood vessel in the body?

Fructose reduces Tollen's reagent due to

Vedclass Test Series

Exam Paper Generator

Online Exam Module