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 μ s. The capacitance of the capacitor is μ F . Given: ln ((10/9))=0.105

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 μ s. The capacitance of the capacitor is μ F . Given: ln ((10/9))=0.105 (A) 9.52 (B) 0.95 (C) 0.105 (D) 1.85

Tardigrade Admin · Accepted Answer

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/R C)=ℓ n((10/9)) ⇒ C=(t/R ℓ n((10)9)) C =(10-6/10 × .105) =.95 μ F

Q. 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 $1 μ s$ . The capacitance of the capacitor is ____ $μ F .$
Given : $ln (\frac{10}{9}) = 0.105$

9.52

0.95

0.105

1.85

$V = V_{0} (1 - e^{- t / RC})$
$2 = 20 (1 - e^{- t / RC})$
$\frac{1}{10} = 1 - e^{- t / RC}$
$e^{- t / RC} = \frac{9}{10}$
$e^{t / RC} = \frac{10}{9}$
$\frac{t}{RC} = ℓ n (\frac{10}{9})$
$\Rightarrow C = \frac{t}{R ℓ n ( \frac{10}{9} )}$
$C = \frac{1 0 ^{- 6}}{10 \times .105}$
$= .95 μ F$

Q. A capacitor is connected to a 20V battery through a resistance of 10Ω. It is found that the potential difference across the capacitor rises to 2V in 1μs. The capacitance of the capacitor is ____ μF. Given : ln(910​)=0.105