Question

A series circuit has a capacitor of 0.25 x 10⁻⁶ F, a resistor of 5 x 10³ ohms, and an inductor of 1 H. The initial charge on the capacitor is zero. If a 12-volt battery is connected to the circuit and the circuit is closed at t=0, determine the charge on the capacitor at t=0.001s, at t=0.01s, and at any time t. Also, determine the limiting charge as t → [infinity]

Answer 1

The charge on the capacitor in a series circuit with a capacitor, resistor, and inductor can be calculated using the formula Q(t) = Q_max * (1 - e^(-t/RC)). At t = 0.001s and t = 0.01s, the charge on the capacitor can be calculated using this formula. At any time t, the charge on the capacitor can be found by substituting the values into the formula. As t approaches infinity, the charge on the capacitor approaches the maximum charge.

Step-by-step explanation:

In a series circuit with a capacitor, resistor, and inductor, the initial charge on the capacitor is zero. To determine the charge on the capacitor at different times, we can use the formula Q(t) = Q_max * (1 - e^(-t/RC)), where Q(t) is the charge at time t, Q_max is the maximum charge (which is equal to the initial charge), R is the resistance, C is the capacitance, and e is Euler's number (approximately 2.71828).

At t = 0.001s, we substitute the values into the formula: Q(0.001) = (0.25 x 10^-6) * (1 - e^(-0.001/(5 x 10^3 * 0.25 x 10^-6))).

At t = 0.01s, we substitute the values into the formula: Q(0.01) = (0.25 x 10^-6) * (1 - e^(-0.01/(5 x 10^3 * 0.25 x 10^-6))).

For any time t, we can substitute the values into the formula to find the charge on the capacitor.

As t approaches infinity, the charge on the capacitor approaches the maximum charge, which is equal to the initial charge on the capacitor.