Two identical capacitors A and B, charged to the same potential 5V are connected in two different circuits as shown below at time t = 0. If the charge on capacitors A and B at time t = CR is QA and QB respectively, then (Here e is the base of natural logarithm)

Question

Two identical capacitors A and B, charged to the same potential 5V are connected in two different circuits as shown below at time t = 0. If the charge on capacitors A and B at time t = CR is QA and QB respectively, then (Here e is the base of natural logarithm) (A) QA=VC, QB=CV (B) QA=(CV/2), QB=(VC/e) (C) QA=(VC/e), QB=(CV/2) (D) QA=VC, QB=(VC/e)

Tardigrade Admin · Accepted Answer

For (A) No current flows Hence QA = CV For (B) i = (V/R)e-(t/RC) q = CV e(t/RC) QB = CVe-1 = (CV/e)

Q. Two identical capacitors $A$ and $B$ , charged to the same potential $5 V$ are connected in two different circuits as shown below at time $t = 0$ . If the charge on capacitors $A$ and $B$ at time $t = CR$ is $Q_{A}$ and $Q_{B}$ respectively, then (Here e is the base of natural logarithm)

$Q_{A} = V C, Q_{B} = C V$

$Q_{A} = \frac{C V}{2}, Q_{B} = \frac{V C}{e}$

$Q_{A} = \frac{V C}{e}, Q_{B} = \frac{C V}{2}$

$Q_{A} = V C, Q_{B} = \frac{V C}{e}$

For (A)
No current flows
Hence $Q_{A} = C V$
For (B)
$i = \frac{V}{R} e^{- \frac{t}{RC}}$
$q = C V e^{\frac{t}{RC}}$
$Q_{B} = C V e^{- 1} = \frac{C V}{e}$

Q. Two identical capacitors A and B, charged to the same potential 5V are connected in two different circuits as shown below at time t=0. If the charge on capacitors A and B at time t=CR is QA​ and QB​ respectively, then (Here e is the base of natural logarithm)

QA​=VC,QB​=CV

QA​=2CV​,QB​=eVC​

QA​=eVC​,QB​=2CV​

QA​=VC,QB​=eVC​