Mastering Physics Solutions

A large capacitor has a charge + q on one plate and – q on the other. At time t=0, the capacitor is connected in series to two ammeters and a light bulb. Immediately after the circuit is closed, the ammeter connected to the positive plate of the capacitor reads IP and the ammeter connected to the negative plate of the capacitor reads IN.

Each ammeter reads positive if current flows through the circuit in a clockwise direction (from the + to the – terminal of the meter).

Immediately after time t=0, what happens to the charge on the capacitor plates?
Check all that apply.

• Individual charges flow through the circuit from the positive to the negative plate of the capacitor.
• Individual charges flow through the circuit from the negative to the positive plate of the capacitor.
• The positive and negative charges attract each other, so they stay in the capacitor.
• Current flows clockwise through the circuit.
• Current flows counterclockwise through the circuit.

Two of the above are correct:

Individual charges flow through the circuit from the negative to the positive plate of the capacitor., Current flows clockwise through the circuit.

At any given instant after t = 0, what is the relationship between the current flowing through the two ammeters, IP and IN, and the current through the bulb, IB?

• IP > IB > IN
• IP = IB > IN
• IP > IB = IN
• IP = IB = IN

IP = IB = IN

What is the relationship between the current and charge? As the charge q(t) on the positive plate of the capacitor decreases, what happens to the value of the current?

The current:

• increases.
• decreases.
• does not change.

decreases.

Light bulbs are often assumed to obey Ohm’s law. However, this is not really true because their resistance increases substantially as the filament heats up in its “working” state.

A typical flashlight bulb at full brilliance draws a current of approximately 0.5 A when connected to a 3-V voltage source. For this problem, assume that the changing resistance causes the current to be 0.5 A for any voltage between 2 and 3 V.

Suppose this flashlight bulb is attached to a capacitor as shown in the circuit from the problem introduction. If the capacitor has a capacitance of 3 F (an unusually large but not unrealistic value) and is initially charged to 3 V, how long will it take for the voltage across the flashlight bulb to drop to 2 V (where the bulb will be orange and dim)? Call this time t_bright.
Express t_bright numerically in seconds to the nearest integer.

The charge on a capacitor Q is equal to C * V. At 3 volts:

Q = CV
Q = 3 * 3
Q = 9

And at 2 volts:

Q = CV
Q = 3 * 2
Q = 6

The difference is 3 coulombs and at a current of 0.5 A (0.5 coulombs per second), it will take about 6 seconds for the voltage to drop.

6 s