Electrostatic-12

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145

Ranadeep Roy ·2013-06-05 05:19:29

83Î¼F

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Akash Anand Excellent ..try to give a crisp solution of this.
Upvote·0· Reply ·2013-06-06 02:54:48
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Dwijaraj Paul Chowdhury ·2013-06-05 04:46:08

8/3 Î¼F

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Akash Anand Excellent work.keep it up
Upvote·0· Reply ·2013-06-06 02:54:20
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1133

Sourish Ghosh ·2013-06-07 07:09:46

Solution please :/

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Shreya Sharma ·2013-06-07 07:19:34

We can join the two points connected by the plain wire as there is no potential drop there. The resulting two 1 Î¼F capacitors are in parallel.
c(parallel)=1+1=2
This is in series with the other 1Î¼F capacitor
1/c = 1 + 1/2 = 3/2
c(series)=2/3
This whole thing is in parallel with the 2Î¼F capacitor
c(total)= 2/3 + 2 = 8/3

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Sourish Ghosh :)
Upvote·0· Reply ·2013-06-07 07:26:45
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Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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Electrostatic-12

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