loading="eager" fetchpriority="high" decoding="sync" /> Beta I Block A, Noida, India - 201308.
5 yrs of Exp
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Beta I Block A, Noida, India - 201308
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Class Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
5
Board
CBSE
Subjects taught
Mathematics, English, Physics, Chemistry
Taught in School or College
No
Upcoming Live Classes
1. Which school boards of Class 12 do you teach for?
CBSE
2. Have you ever taught in any School or College?
No
3. Which classes do you teach?
I teach Class 12 Tuition Class.
4. Do you provide a demo class?
Yes, I provide a free demo class.
5. How many years of experience do you have?
I have been teaching for 5 years.
Answered on 18/10/2019 Learn CBSE - Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
Here, C₁ = 600pF , V₁ = 20pV , C₂ = 600pF and V₂ = 0
On connecting charged capacitor to uncharged capacitor , the common potential V across the capacitors is given by
V = (C₁V₁ + C₂V₂)/(C₁ + C₂)
= (600 × 10⁻¹² × 200 + 600 × 10⁻¹² × 0)/(600 + 600) × 10⁻¹²
= 100V
now, energy stored in capacitors before connection is
Ui = 1/2 C₁V₁² + 1/2 C₂V₂²
= 1/2 × 600 × 10⁻¹² × (200)² + 1/2 × 600 × 10⁻¹² × 0
= 12 × 10⁻⁶ J = 12μJ
And energy stored in capacitors after connection is
Uf = 1/2 (C₁ + C₂)V²
= 1/2 × (600 + 600) × 10⁻¹² × (100)²
= 1/2 × 1200 × 10⁻⁸ J
= 6 × 10⁻⁶ J
= 6μJ
Hence , energy lost in the process is ∆U = Uf - Ui
= 6μJ - 12μJ = -6μJ [ negative sign indicates energy is lost.
Answered on 12/10/2019 Learn CBSE - Class 12/Science/Chemistry/Unit IX: Coordination Compounds/NCERT Solutions/Intext Exercise 9.1
Class Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
5
Board
CBSE
Subjects taught
Mathematics, English, Physics, Chemistry
Taught in School or College
No
Answered on 18/10/2019 Learn CBSE - Class 12/Science/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
Here, C₁ = 600pF , V₁ = 20pV , C₂ = 600pF and V₂ = 0
On connecting charged capacitor to uncharged capacitor , the common potential V across the capacitors is given by
V = (C₁V₁ + C₂V₂)/(C₁ + C₂)
= (600 × 10⁻¹² × 200 + 600 × 10⁻¹² × 0)/(600 + 600) × 10⁻¹²
= 100V
now, energy stored in capacitors before connection is
Ui = 1/2 C₁V₁² + 1/2 C₂V₂²
= 1/2 × 600 × 10⁻¹² × (200)² + 1/2 × 600 × 10⁻¹² × 0
= 12 × 10⁻⁶ J = 12μJ
And energy stored in capacitors after connection is
Uf = 1/2 (C₁ + C₂)V²
= 1/2 × (600 + 600) × 10⁻¹² × (100)²
= 1/2 × 1200 × 10⁻⁸ J
= 6 × 10⁻⁶ J
= 6μJ
Hence , energy lost in the process is ∆U = Uf - Ui
= 6μJ - 12μJ = -6μJ [ negative sign indicates energy is lost.
Answered on 12/10/2019 Learn CBSE - Class 12/Science/Chemistry/Unit IX: Coordination Compounds/NCERT Solutions/Intext Exercise 9.1
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Beta I Block A, Noida 
