If the condenser shown in the circuit is charged to 5 V and left in the circuit, in 12 s the charge on the condenser will become:

Question

If the condenser shown in the circuit is charged to 5 V and left in the circuit, in 12 s the charge on the condenser will become: (A)  (10/e) coulomb (B)  (e2/10) coulomb (C)  (10/e2) coulomb (D)  (e/10) coulomb

Tardigrade Admin · Accepted Answer

Exponential decay of charge takes place.
In a C-R circuit, discharging takes place.
The

q - t

equation is

q = q_{0} e^{- t / RC}

where

R

is resistance,

C

is capacitance
and

q_{0} = C V_{0}

Hence,

q = C V_{0} e^{- t / RC}

Given,

V_{0} = 5 V, C = 2 F, t = 12 s, R = 6 Ω

∴ q = 2 \times 5 e^{- 12/6 \times 2} = \frac{10}{e}

coulomb

Q. If the condenser shown in the circuit is charged to $5 V$ and left in the circuit, in $12 s$ the charge on the condenser will become:

$\frac{10}{e}$ coulomb

$\frac{e ^{2}}{10}$ coulomb

$\frac{10}{e ^{2}}$ coulomb

$\frac{e}{10}$ coulomb

Q. If the condenser shown in the circuit is charged to 5V and left in the circuit, in 12s the charge on the condenser will become:

e10​ coulomb

10e2​ coulomb

e210​ coulomb

10e​ coulomb

Exponential decay of charge takes place. In a C-R circuit, discharging takes place. The q−t equation is q=q0​e−t/RC where R is resistance, C is capacitance and q0​=CV0​ Hence, q=CV0​e−t/RC Given, V0​=5V,C=2F,t=12s,R=6Ω ∴q=2×5e−12/6×2=e10​ coulomb

Q. If the condenser shown in the circuit is charged to $5 V$ and left in the circuit, in $12 s$ the charge on the condenser will become:

$\frac{10}{e}$ coulomb

$\frac{e ^{2}}{10}$ coulomb

$\frac{10}{e ^{2}}$ coulomb

$\frac{e}{10}$ coulomb