Ask question

Ask question

All categories

3 years ago

11

Why is it called field effect transistor

1 answer:

Mrac [35]3 years ago

5 0

Answer:

The field-effect transistor (FET) is a type of transistor which uses an electric field to control the flow of current.

FET is named field effect transistor as field effect is producing, but in BJT also field effect will produce. :)

You might be interested in

Explain mathematically whether the divergence of the curl of a vector field is

svp [43]

Answer:

yes

Explanation:

6 0

2 years ago

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f

Gelneren [198K]

Answer:

(a) $Q=332$ $kvar$ and $C=5.66$ $uF$

(b) $pf=0.90$ lagging

Explanation:

Given Data:

$P=600kW$

$V=12.47kV$

$f=60Hz$

$pf_{old} =0.75$

$pf_{new} =0.95$

(a) Find the required kVAR rating of a capacitor

$\alpha _{old}=cos^{-1}(0.75) =41.41$ °

$\alpha _{new}=cos^{-1}(0.95) =18.19$ °

The required compensation reactive power can be found by

$Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))$

$Q=600(tan(41.41) - tan(18.19))$

$Q=332$ $kvar$

The corresponding capacitor value can be found by

$C=Q/2\pi fV^{2}$

$C=332/2*\pi *60*12.47^{2}$

$C=5.66$ $uF$

(b) calculate the resultant supply power factor

First convert the hp into kW

$P_{mech} =250*746=186.5$ $kW$

Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

$I_{m} =(233.125/12.47*0.85)$

$I_{m}=18.694+11.586$ $j$ $A$

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

$I_{load} =(600/12.47*0.75)$

$I_{load} =48.115-42.433$ $j$ $A$

Now the supply current is the current flowing in the load plus the current flowing in the motor

$I_{supply} =I_{m} + I_{load}$

$I_{supply}= (18.694+11.586)+(48.115-42.433)$

$I_{supply} =66.809-30.847j$ $A$

or in polar form

$I_{supply} =73.58$ °

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

$pf=cos(24.78)=0.90$ lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0

3 years ago

Unfiltered full wave rectifier with a 120 V 60 Hz input produces an output with a peak of 15V. When a capacitor-input filter and

Alborosie

Answer:

$V_{pp}=2V$

Explanation:

Source Voltage $V= 120V$

Frequency $f=60Hz$

Peak output voltage $Vp=15V$

Peak Output Voltage with filter $V_p'=14V$

Generally the equation for Peak to peak voltage is mathematically given by

$V_p'=V_p-\frac{V_{pp}}{2}$

Therefore

$V_{pp}=2(V_p-v_p')$

$V_{pp}=2(15-14)$

$V_{pp}=2V$

5 0

2 years ago

. Analyze the following scenarios and mention in which normal form it is with reason? a) Book- bookID bookTitle PublisherName Pu

svet-max [94.6K]

Answer:

You so good is ué KKKKKK a gente não não é possível que o seu ponto de vista da minha irmã me ligou

Explanation:

Irieiwy Wwowuwowywbrbbdnd

8 0

3 years ago

A man can swim at 4 ft/s in still water. He wishes to cross tje 40-ft wide river to point B, 30 ft downstream. If the river flow

iVinArrow [24]

Answer:

$v \approx 4.472\,\frac{ft}{s}$ , $t = 10\,s$

Explanation:

Since man and river report constant speeds and velocities are mutually perpendicullar, the absolute speed of the man is calculated by the Pythagorean Theorem:

$v = \sqrt{(4\,\frac{ft}{s} )^{2}+(2\,\frac{ft}{s} )^{2}}$

$v \approx 4.472\,\frac{ft}{s}$

The required time to make the crossing is:

$t = \frac{40\,ft}{4\,\frac{ft}{s} }$

$t = 10\,s$

6 0

3 years ago