A series RLC circuit is connected to a a DC battery via a switch.
#11A (3 points possible)
A long time after the switch is closed, what is the current in the circuit?
A long time after the switch is closed, what is the current in the circuit? - unanswered
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#11B (3 points possible)
A long time after the switch is closed, over which element does the potential created by the battery drop?
the switch the switch - incorrect the resistor the capacitor the inductor
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A long time after the switch is closed, the current in the circuit would be zero. This is because an RLC circuit is a series circuit with an inductor, capacitor, and resistor. When a DC battery is connected to such a circuit, the inductor initially resists any change in current, causing the current to increase slowly. However, over time, the inductor's magnetic field becomes stable, and the current becomes constant. So, after a long time, the current in the circuit would be zero.
A long time after the switch is closed, the potential created by the battery drops primarily over the resistor. In a series circuit, the total voltage across all elements is equal to the battery voltage. Since the inductor resists changes in current and the capacitor acts as an open circuit for DC, the resistor is the primary element over which the potential drops in the circuit.
To answer these questions, we need to understand the behavior of a series RLC circuit connected to a DC battery.
A series RLC circuit consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series. When a direct current (DC) source, such as a battery, is connected to the circuit via a switch, various behaviors occur.
#11A: A long time after the switch is closed, what is the current in the circuit?
To determine the long-term behavior of the circuit after the switch is closed, we need to consider the inductor and capacitor's properties.
An inductor opposes changes in current, so it tends to keep the current flowing in the circuit relatively constant.
A capacitor, on the other hand, stores electrical charge and opposes changes in voltage. In the long run, a fully charged capacitor acts as an open circuit, blocking current flow.
Since we're dealing with a DC battery, which has a constant voltage, the inductor and capacitor will eventually reach their final states.
In a long time after the switch is closed, the current in the circuit will be determined by the resistance (R) alone. The inductor and capacitor will have no impact on the current flow. Therefore, the current in the circuit will be determined solely by Ohm's Law: I = V/R, where V is the voltage provided by the battery and R is the resistance in the circuit.
#11B: A long time after the switch is closed, over which element does the potential created by the battery drop?
In a long time after the switch is closed, the potential created by the battery drops only across the resistor (R). This is because the inductor eventually reaches a steady state where it resists current changes, acting as an open circuit, and the capacitor is fully charged, also acting as an open circuit.
In summary, a long time after the switch is closed, the current in the circuit will be determined by the resistance (R) alone, while the potential created by the battery will drop entirely across the resistor (R) only.