A capacitor of capacitance C = 1µF is suddenly connected to a battery of 100 volt through a resistance R = 100Ω. The time taken for the capacitor to be charged to get 50 V is: [Take ln 2 = 0.69]

Option 1 - 3.33 × 10⁻⁴ s

Option 2 - 1.44 × 10⁻⁴ s

Option 3 - 0.30 × 10⁻⁴ s

Option 4 - 0.69 × 10⁻⁴ s

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Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

Correct Option - 4

Detailed Solution:

V = E (1 - e? /τ)
Where τ = RC = 100 * 10? = 10? sec
50V = 100V (1 - e? /10? )
1/2 = 1 - e? ¹?
1/2 = e? ¹?
-ln2 = -10? t
t = ln2/10?
t = 0.693 * 10? sec

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