The amplitude of the charge oscillating in a circuit decreases exponentially as Q=Q_0 e^ -R t 2 L , where Q_0 is the charge at t=0 ~s . The time at which charge amplitude decreases to 0.50 Q_0 is nearly: [Given that

Correct answer: B. 11.09 ms Given Q=Q_0 e^ -R t / 2 L R=1.5 , L=12 mH, In(2)=0.693 Q=0.5 Q_0, t= ? 0.5 Q_0=Q_0 e^ -R t / 2 L (1)/(2)=e^ -R t / 2 L Taking log on both sides ((1)/(2))= e^ -R t / 2 L 2=(R t)/(2 L) t= 2 L In 2 R = 2 × 12 × 10^ -3 × 0.693 1.5 t=11.09 ~ms

NEET 2024

Exam: NEET
Year: 2024
Subject: Physics
Chapter: Alternating Current

NEET 2024 · Physics · Alternating Current

The amplitude of the charge oscillating in a circuit decreases...

The amplitude of the charge oscillating in a circuit decreases exponentially as

Q=Q_0 e^{-R t 2 L}

, where

Q_0

is the charge at

t=0 \mathrm{~s}

. The time at which charge amplitude decreases to

0.50 Q_0

is nearly: [Given that

R=1.5 \Omega, L=12 \mathrm{mH}, \ln (2)=0.693

]

Options

A19.01 ms
B11.09 ms
C19.01 s
D11.09 s

Correct answer

B. 11.09 ms

Step-by-step solution

Given

Q=Q_0 e^{-R t / 2 L}

R=1.5 \Omega, L=12 \mathrm{mH}, \operatorname{In}(2)=0.693

Q=0.5 Q_0, t=

0.5 Q_0=Q_0 e^{-R t / 2 L}

\Rightarrow \quad \frac{1}{2}=e^{-R t / 2 L}

Taking log on both sides

\ln \left(\frac{1}{2}\right)=\ln e^{-R t / 2 L}

\Rightarrow \ln 2=\frac{R t}{2 L}

t=\frac{2 L \operatorname{In} 2}{R}=\frac{2 \times 12 \times 10^{-3} \times 0.693}{1.5}

t=11.09 \mathrm{~ms}

Practice Alternating Current on Quantrex Academy →

More from Alternating Current