Question 441: AC fundamentals

Question

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

Accepted Answer

The given equation for the amplitude of the charge oscillating in a circuit is:
$$Q = Q_0 e^{- \frac{R t}{2 L} }$$
We need to find the time, $$t$$, at which charge amplitude decreases to $$0.50 Q_0$$. So, we set $$Q = 0.50 Q_0$$:
$$0.50 Q_0 = Q_0 e^{- \frac{R t}{2 L} }$$
Divide both sides by $$Q_0$$:
$$0.50 = e^{- \frac{R t}{2 L} }$$
Take the natural logarithm on both sides to solve for $$t$$:
$$\ln(0.50) = - \frac{R t}{2 L}$$
Recall that $$\ln(0.50) = - \ln(2)$$. Substituting the given value $$\ln(2) = 0.693$$, we get:
$$-0.693 = - \frac{R t}{2 L}$$
Remove the negative signs from both sides:
$$0.693 = \frac{R t}{2 L}$$
Now, solve for $$t$$:
$$t = \frac{2 L \cdot 0.693}{R}$$
Substituting the given values $$R = 1.5 \Omega$$ and $$L = 12 \mathrm{~mH} = 12 	imes 10^{-3} \mathrm{~H}$$, we have:
$$t = \frac{2 	imes 12 	imes 10^{-3} \cdot 0.693}{1.5}$$
Calculate the numerator and denominator:
$$t = \frac{16.632 	imes 10^{-3}}{1.5}$$
Finally, compute the value of $$t$$:
$$t = 11.09 	imes 10^{-3} \mathrm{~s} = 11.09 \mathrm{~ms}$$
Therefore, the time at which the charge amplitude decreases to $$0.50 Q_0$$ is nearly 11.09 ms. Thus, the correct answer is:
Option B: 11.09 ms

Answer

19.01 ms

Answer

11.09 ms

Answer

19.01 s

Answer

11.09 s

AC fundamentals

The amplitude of the charge oscillating in a circuit decreases exponentially as $Q=Q_0 e^{-R t/2 L}, where Q_0 is the charge at t=0 \mathrm{~s}. The time at which charge amplitude decreases to 0.50 Q_0 is nearly: [Given that R=1.5 \Omega, L=12 \mathrm{~mH}, \ln (2)=0.693$]

Choose the correct answer: