Electric Current - AP Physics 1 Exam Question

A circuit consists of a 3 ohm resistor and a 5 farad capacitor connected in series with a 12 volt battery. The switch is closed at t = 0 seconds.

(a) Determine the magnitude and direction of the initial current flowing through the circuit immediately after the switch is closed.

(b) If the circuit has a total resistance of 6 ohms, calculate the time constant of the circuit.

(c) Sketch a graph representing the charge on the capacitor as a function of time.

(d) Calculate the magnitude and direction of the current through the circuit at t = 2 seconds.

Answer:

(a) To determine the magnitude and direction of the initial current, we can use Ohm's law. Ohm's law states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R). In this case, we have a single resistor.

Using Ohm's law, we can calculate the initial current (I) as:

I = V / R

Plugging in the given values, we have:

I = 12 V / 3 Ω

Simplifying, we find:

I = 4 A

Since the battery is providing a voltage of 12 volts, the current will flow from the positive terminal of the battery to the negative terminal, in the direction opposite to the conventional current flow (from positive to negative).

Therefore, the initial current has a magnitude of 4 Amperes and is directed from the negative terminal of the battery towards the positive terminal.

(b) The time constant (τ) of a circuit in an RC circuit is given by the formula τ = RC, where R is the total resistance and C is the capacitance. In this case, we have a total resistance of 6 ohms.

Using the given values, we find:

τ = (6 Ω)(5 F)

Simplifying, we have:

τ = 30 s

Therefore, the time constant of the circuit is 30 seconds.

(c) The charge on a capacitor in an RC circuit is given by the equation Q(t) = Q₀(1 - e^(-t/τ)), where Q₀ is the initial charge on the capacitor, t is the time, and τ is the time constant.

In this case, the graph of charge on the capacitor as a function of time will start at zero, gradually increase, and approach a final value. The rate at which the charge increases depends on the time constant of the circuit.

(d) To calculate the magnitude and direction of the current at t = 2 seconds, we can use the equation for charge on the capacitor and differentiate it with respect to time to find the current.

Differentiating Q(t) = Q₀(1 - e^(-t/τ)) with respect to time, we get:

I(t) = (dQ/dt) = (Q₀/τ) * e^(-t/τ)

Plugging in t = 2 seconds and τ = 30 seconds, we have:

I(2) = (Q₀/30) * e^(-2/30)

To calculate the magnitude of the current, we need to know the initial charge on the capacitor (Q₀). However, since the initial charge is not provided in the question, we cannot determine the exact magnitude of the current at t = 2 seconds. The direction of the current will be determined by the sign of I(2), which can only be determined if we know the initial charge.

Therefore, we cannot calculate the magnitude and direction of the current at t = 2 seconds without knowing the initial charge on the capacitor.