All categories

2 years ago

1. A silicon BJT is connected as shown in Fig 1, where RC = 3.6 k 2. VBE = 0.8 V. (10%)

Physics

1 answer:

Masja [62]2 years ago

4 0

Answer:

The circuit is missing attached below is the required circuit

answer :

a) Ic = 1.944 mA

Rp = 288.66 kΩ

b) The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased

Explanation:

Rc = 3.6 kΩ

VBE = 0.8 v

1) predict Ic and specify Rp to establish Vce at 5 V

we will apply Kirchhoff's voltage law to resolve this

solution attached below

b ) The BJT is said to be in Forward reverse bias because The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased

You might be interested in

True or False:In a current, electrons will always flow from negative to positive.

DIA [1.3K]

It should be true: electrons low from negative toward positive to negative toward positive because opposite charges attract each other.

I hope this was correct

3 0

3 years ago

Read 2 more answers

A food web illustrates a snake eating a rat. What is the role of the snake in the food web? A. Decomposer B.Secondary consumer C

Svetradugi [14.3K]

Primary Consumer or C.

5 0

3 years ago

Read 2 more answers

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

3 years ago

A wooden block has a mass of 562 g and a volume of 72 cm3. What is the density?

dangina [55]

Mass/volume is density so it’s 562g/72cm^3 so it’s roughly 7.805g per cubic centimeter

3 0

3 years ago

Read 2 more answers

On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze

sp2606 [1]

Answer:

0.78333 m/s in the opposite direction

1.566 m/s in the same direction

Explanation:

$m_1$ = Mass of penny = 0.0025 kg

$m_2$ = Mass of nickel = 0.005 kg

$u_1$ = Initial Velocity of penny = 2.35 m/s

$u_2$ = Initial Velocity of nickel = 0 m/s

$v_1$ = Final Velocity of penny

$v_2$ = Final Velocity of nickel

As momentum and Energy is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

${\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}$

From the two equations we get

$v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.0025-0.005}{0.0025+0.005}\times 2.35+\frac{2\times 0.5}{0.4005+0.5}\times 0\\\Rightarrow v_1=-0.78333\ m/s$

The final velocity of the penny is 0.78333 m/s in the opposite direction

$v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.0025}{0.0025+0.005}\times 2.35+\frac{0.005-0.0025}{0.005+0.0025}\times 0\\\Rightarrow v_2=1.566\ m/s$

The final velocity of the nickel is 1.566 m/s in the same direction

6 0

2 years ago