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]

3.33 × 10⁻⁴ s

1.44 × 10⁻⁴ s

0.30 × 10⁻⁴ s

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Electrostatic Potential Physics Electrostatic Potential and Capacitance Physics Class 12th

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

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month 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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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]

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