Question

A capacitor is connected to a 20 V battery through a resistance of 10 Ω. It is found that the potential difference across the capacitor rises to 2 V in 1 µs. The capacitance of the capacitor is ______ µF. Given: In `(10/9)` = 0.105

पर्याय

• 9.52

• 0.95

• 0.105

• 1.85

Answer 1

A capacitor is connected to a 20 V battery through a resistance of 10 Ω. It is found that the potential difference across the capacitor rises to 2 V in 1 µs. The capacitance of the capacitor is 0.95 µF. Given: In `(10/9)` = 0.105

Explanation:

Given,

Voltage of battery, V₀ = 20V 

Voltage across capacitor, V = 2V

V = `"V"_0 (1-"e"^(-"t"//"RC"))=> 2 = 20(1-"e"^(-"t"//"RC"))` 

⇒ `1/10=1-e^(-"t"//"RC") =>"e"^(-"t"//"RC") = 9/10 => "e"^("t"//"RC") = 10/9`

⇒ `"t"/"RC" =ℓ"n"(10/9)=>"C"= "t"/("R"ℓ"n"(10/9))`

⇒ C = `10^-6/(10xx0.105)`

= 0.95 µF.