First slide

Diodes

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 $t = C R$ is $Q_{A} a n d Q_{B}$ respectively, then (here e is the base of natural logarithm)

Moderate

A

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

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

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

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

Solution

For A
Diode is in reverse bias condition
No current flows $\Rightarrow Q_{A} = C V$

For B

Capacitor is discharging
$q = q_{0} e^{- \frac{t}{c R}}$
$q = C V . e^{- \frac{t}{c R}}$
At t=CR
$q = C V . e^{- 1} = \frac{C V}{e}$

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