Illumination Engineering Part 1 | Utilization of Electrical Energy | Nikhil Nakka

[@sebastian6405]

Hello sir this video is very helpful for us btw your Mustache looking so cool

[@varalakshmikorrapati7218]

Can u provide the pdf for illumination and traction ........plzzz

[@sekharvadde1141]

Sir your explanation is awesome....and easy to understand the subject...tq

[@csrivishnureddy]

Heat is not a wave its the internal energy of matter (mostly vibrational energy of atoms) and infrared radiation is emitted by any matter that is above absolute temp.(-273 C) ........

[@anjalirani1732]

Please sir upload 4-1 subject utilisation of electrical energy all topic

[@noobsteamgaming8436]

I want

[@nitya7670]

Incredible.......!!!!!!!!!

[@ashokgoud56]

Sir,Tsspdcl ni private chesthadanta gaa nijamena .?

[@priyankkayasth6661]

Just superb.... ☺☺

[@amitkumarsoni3310]

Next part sir pls

[@Practic3_gratitud3]

Very nice explanation sir

[@vickyvikashmalhotra8246]

Sir iska next part ka link send kijiye plz

[@abhishekbhardwaj2382]

Where are the other series of these illumination ?

[@nSackStyles]

Could you please provide me the pdf of "Utilization of Electrical Energy" by JB Gupta. My father needs it to deliver the lectures.

[@cartoonsforkids8105]

sir where is its 2 part

[@parisnarasimha8868]

It is required to provide an illumination of 100 lm/m2 in a workshop hall 40m × 10m and efficiency of lamp is 14 lm/W. Calculate the number and rating of lamps and their positions when trusses are provided at mutual distance of 5m. Take coefficient of utilisation as 0.4 and depreciation factor as 0.8. Sir, Can you solve this problem.. thanks you

[@archanam7412]

pls take this class on unacademy under ssc exams

[@s2-gatemeditatn273]

Great video👌 where is next part ?

[@shivamrvk]

Superb........

[@sateeshtechhub5908]

It is F sir. Not Q. Light is denoted by Q.

[@avneeshgupta7918]

Very good

[@sateeshkumar6089]

Sir where is the power system generation part 2 to 6? Can you provide me the link....

[@princejoseph1349]

part 2 where is it?

[@nikhilnakka]

today night

[@ronieu1]

Sir, may I ask you a question how to mature the external quantum efficiency if A particular deep blue commercial LED emits an optical power of453mW at 455 nm when the current is 350 mA and the forward voltage is 3.2 V? Thank you.

[@Pavankalyanreddy-q3t]

To determine the external quantum efficiency (EQE) of a deep blue LED, you need to follow these steps: Calculate the input electrical power: The input electrical power 𝑃 𝑖 𝑛 P in ​ is given by the product of the current 𝐼 I and the forward voltage 𝑉 V: 𝑃 𝑖 𝑛 = 𝐼 × 𝑉 P in ​ =I×V Given: 𝐼 = 350   mA = 0.35   A I=350mA=0.35A 𝑉 = 3.2   V V=3.2V Therefore: 𝑃 𝑖 𝑛 = 0.35   A × 3.2   V = 1.12   W P in ​ =0.35A×3.2V=1.12W Calculate the output optical power: Given the output optical power 𝑃 𝑜 𝑝 𝑡 P opt ​ is 453 mW, convert it to watts: 𝑃 𝑜 𝑝 𝑡 = 453   mW = 0.453   W P opt ​ =453mW=0.453W Calculate the photon energy: The energy of a single photon 𝐸 E can be calculated using the wavelength 𝜆 λ and the speed of light 𝑐 c as well as Planck's constant ℎ h: 𝐸 = ℎ × 𝑐 𝜆 E= λ h×c ​ Where: ℎ = 6.626 × 1 0 − 34   J ⋅ s ( Planck’s constant ) h=6.626×10 −34 J⋅s(Planck’s constant) 𝑐 = 3 × 1 0 8   m/s ( speed of light ) c=3×10 8 m/s(speed of light) 𝜆 = 455   nm = 455 × 1 0 − 9   m λ=455nm=455×10 −9 m Therefore: 𝐸 = 6.626 × 1 0 − 34 × 3 × 1 0 8 455 × 1 0 − 9 ≈ 4.37 × 1 0 − 19   J E= 455×10 −9 6.626×10 −34 ×3×10 8 ​ ≈4.37×10 −19 J Calculate the total number of photons emitted per second: The optical power 𝑃 𝑜 𝑝 𝑡 P opt ​ is related to the number of photons 𝑁 N emitted per second: 𝑃 𝑜 𝑝 𝑡 = 𝑁 × 𝐸 P opt ​ =N×E Therefore: 𝑁 = 𝑃 𝑜 𝑝 𝑡 𝐸 = 0.453 4.37 × 1 0 − 19 ≈ 1.04 × 1 0 18   photons/s N= E P opt ​ ​ = 4.37×10 −19 0.453 ​ ≈1.04×10 18 photons/s Calculate the EQE: The external quantum efficiency (EQE) is the ratio of the number of photons emitted to the number of electrons passing through the LED. The number of electrons per second is given by the current 𝐼 I divided by the elementary charge 𝑒 e: Number of electrons per second = 𝐼 𝑒 = 0.35 1.602 × 1 0 − 19 ≈ 2.18 × 1 0 18   electrons/s Number of electrons per second= e I ​ = 1.602×10 −19 0.35 ​ ≈2.18×10 18 electrons/s Therefore, the EQE is: EQE = Number of photons emitted per second Number of electrons per second = 1.04 × 1 0 18 2.18 × 1 0 18 ≈ 0.477   ( or 47.7 % ) EQE= Number of electrons per second Number of photons emitted per second ​ = 2.18×10 18 1.04×10 18 ​ ≈0.477(or 47.7%) Thus, the external quantum efficiency (EQE) of the deep blue LED is approximately 47.7%.

[@tianahtina5486]

thank you sir for this great lesson i have leart may God bless u please

[@nikhilnakka]

Thank you so much! Do subscribe my channel and checkout other good videos too!!

[@gnaveen8451]

Class supar

[@manivannans9928]

Bro network theory test solution vedio is enough

[@gauravnikki3577]

Sir kya Hindi me ni padha sk te h

[@ranaradhaofficial]

Please sir hindi me samjhayiye......

[@dineshsinhpuwar5612]

Speak in gujarati

[@keshavsharma-vd4qg]

Hindi me bolo shab

[@pathulothusharada7244]

Hiii nikhil iam sharada

[@kamalshankerpotter4982]

Speaking Hindi