You visited us 1 times! Enjoying our articles? Unlock Full Access!

V-I Characteristics for a P-N Junction Diode

A capacitor ...

Question

A capacitor $(C = 4.0 μ F)$ is connected through a resistor $(R = 2.5 M Ω)$ across a battery of negligible intemal resistance of voltage 12 volts. The time after which the potential difference across the capacitor becomes three times to that of resistor is $(ln 2 = 0.693)$

13.86 s e c .

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6.48 s e c .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.24 s e c .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20.52 s e c .

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $13.86 s e c .$
The charging equation in CR circuit is given by $q = C V_{0} (1 - e^{- t / C R})$

and current $i = \frac{d q}{d t} = \frac{V_{0}}{R} e^{- t / C R}$

here, $V_{C} = 3 V_{R}$

or $V_{0} (1 - e^{- t / C R}) = 3 i R = 3 R \frac{V_{0}}{R} e^{- t / C R}$

or $e^{- t / 10} = 1 / 4$ where $C R = (4 \times 10^{- 6}) \times (2.5 \times 10^{6}) = 10$

or $e^{t / 10} = 4$

taking ln both sides, $t / 10 = l n 4 = 2 l n 2 \Rightarrow t = 20 l n 2 = 13.86 s$

Suggest Corrections

Similar questions

4 μ F

2.5 M Ω

12 V

3

l n (b) = 0.693]

4 μ F

12 v o l t

2.5 M Ω

V_{C} = 3 V_{R}

l n 2 = 0.693

1 μ F

12 V

t = \frac{1}{10} sec

6 V

R

ln 2 = 0.693

15 V

10 Ω

2 μ s,

5 V

(ln 1.5 = 0.4)

L = 400 m H

R_{1} = 2 Ω

R_{2} = 2 Ω

12

V

S

t = 0

L

Join BYJU'S Learning Program