At time , a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V? [Take: ] [2010]

Question

At time $t = 0$ , a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V [Take: $l n 5 = 1.6, l n 3 = 1.1$ ] [2010]

Smart Achievers · Accepted Answer

(2)

The equivalent circuit of the given circuit is as shown below.

$R = 1 M Ω,$ $C = 4 μ F$

$∴$ The time constant

$τ = R C = 4 s$

Potential across $4 μ F$ capacitor at any time $' t'$

$V = V_{0} [1 - e^{- t / τ}]$

$\Rightarrow 4 = 10 [1 - e^{- t / 4}]$

$\Rightarrow t = 2 s \Rightarrow e^{- t / 4} = 0.6 = \frac{3}{5}$

Taking logarithm on both sides,

$- \frac{t}{4} = \ln 3 - \ln 5$ $\Rightarrow t = 4 (\ln 5 - \ln 3) \approx 2.04 s$

Solution

Q.
At time $t = 0$ , a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V [Take: $l n 5 = 1.6, l n 3 = 1.1$ ] [2010]

Want Us to Email you About Special Offers & Updates?

Solution

Q. At time t=0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V [Take: ln5=1.6, ln3=1.1] [2010]

Ans. (2) The equivalent circuit of the given circuit is as shown below. R=1 MΩ, C=4 μF ∴ The time constant τ=RC=4 s Potential across 4 μF capacitor at any time 't' V=V0[1-e-t/τ] ⇒4=10[1-e-t/4] ⇒t=2 s⇒e-t/4=0.6=35 Taking logarithm on both sides, -t4=ln3-ln5⇒t=4(ln5-ln3)≈2.04 s

Want Us to Email you About Special Offers & Updates?

Q.
At time $t = 0$ , a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V [Take: $l n 5 = 1.6, l n 3 = 1.1$ ] [2010]