Question

In the circuit shown the capacitor and inductor do not have energy initially. Then

• A. the current flowing through the battery at t = 0 is $\epsilon /2R$
• B. the current flowing through the battery at t = 0 is $2\epsilon /3R$
• C. the current flowing through the battery at t = $\infty$ is $2\epsilon /5R$
• D. the current flowing through the battery at t = $\infty$ is $\epsilon /2R$

Answer 1

Answer: (A), (C)

the current flowing through the battery at t = 0 is $\epsilon /2R$
the current flowing through the battery at t = $\infty$ is $2\epsilon /5R$

At the exact instant when power is applied, the capacitor has 0V of stored voltage and so it consumes a theoretically infinite current limited by the series resistance. (A short circuit) As time continues and the charge accumulates, the capacitor's voltage rises and it's current consumption drops until the capacitor voltage and the applied voltage are equal and no current flows into the capacitor (open circuit). For an inductor, the opposite is true. At the moment of power-on, when voltage is first applied, it has a very high resistance to the changed voltage and carries little current (open circuit), and after some time, it will have a low resistance to the steady voltage and carry lots of current (short circuit).
Current at $t=0$ and $t=\infty$ is $\frac{ϵ}{2R}$