Question

Consider the circuit below. The capacitor has a capacitance of 19 mF and starts out uncharged. The switch is closed at time t = 0. [Diagram: Bridge circuit with capacitor C in the middle, battery V = 12 V, switch on top left. Resistors: R1 = 2 ? (upper left), R3 = 3 ? (upper right), R2 = 4 ? (lower left), R4 = 3 ? (lower right).] a. At time t = 0 (immediately after the switch is closed), what is the current flowing out from the battery? Hint for (a) At the moment an uncharged capacitor is connected, it acts as a short, meaning it could be replaced by a wire, for that moment in time. How does this allow you to simplify the remaining resistor-network circuit? At time t = 0, the current from the battery is ___ A. b. A long time after the switch is closed (and capacitor is fully charged), what is the current flowing out from the battery? Hint for (b) After the switch has been closed for a very long time, the capacitor acts as an open circuit, meaning a broken connection across the capacitor. How does this allow you to simplify the remaining resistor-network circuit? A long time after the switch is closed, the current from the battery is _ A. c. What is the charge on the capacitor a long time after the switch is closed? Hint for (c) With the simplifications in (b), you should be able to quickly find the current values, which will allow you to calculate the voltages at the two ends of the charged capacitor. The voltage difference will give you the amount of charge on the fully-charged capacitor. The charge on the capacitor a long time after the switch is closed is _____ mC.

          Consider the circuit below. The capacitor has a capacitance of 19 mF and starts out uncharged. The switch is closed at time t = 0.

[Diagram: Bridge circuit with capacitor C in the middle, battery V = 12 V, switch on top left. Resistors: R1 = 2 ? (upper left), R3 = 3 ? (upper right), R2 = 4 ? (lower left), R4 = 3 ? (lower right).]

a. At time t = 0 (immediately after the switch is closed), what is the current flowing out from the battery?

Hint for (a)
At the moment an uncharged capacitor is connected, it acts as a short, meaning it could be replaced by a wire, for that moment in time. How does this allow you to simplify the remaining resistor-network circuit?

At time t = 0, the current from the battery is _______ A.

b. A long time after the switch is closed (and capacitor is fully charged), what is the current flowing out from the battery?

Hint for (b)
After the switch has been closed for a very long time, the capacitor acts as an open circuit, meaning a broken connection across the capacitor. How does this allow you to simplify the remaining resistor-network circuit?

A long time after the switch is closed, the current from the battery is _______ A.

c. What is the charge on the capacitor a long time after the switch is closed?

Hint for (c)
With the simplifications in (b), you should be able to quickly find the current values, which will allow you to calculate the voltages at the two ends of the charged capacitor. The voltage difference will give you the amount of charge on the fully-charged capacitor.

The charge on the capacitor a long time after the switch is closed is _______ mC.

Consider the circuit below. The capacitor has a capacitance of 19 mF and starts out uncharged. The switch is closed at time t = 0.

[Diagram: Bridge circuit with capacitor C in the middle, battery V = 12 V, switch on top left. Resistors: R1 = 2 ? (upper left), R3 = 3 ? (upper right), R2 = 4 ? (lower left), R4 = 3 ? (lower right).]

a. At time t = 0 (immediately after the switch is closed), what is the current flowing out from the battery?

Hint for (a)
At the moment an uncharged capacitor is connected, it acts as a short, meaning it could be replaced by a wire, for that moment in time. How does this allow you to simplify the remaining resistor-network circuit?

At time t = 0, the current from the battery is A.

b. A long time after the switch is closed (and capacitor is fully charged), what is the current flowing out from the battery?

Hint for (b)
After the switch has been closed for a very long time, the capacitor acts as an open circuit, meaning a broken connection across the capacitor. How does this allow you to simplify the remaining resistor-network circuit?

A long time after the switch is closed, the current from the battery is A.

c. What is the charge on the capacitor a long time after the switch is closed?

Hint for (c)
With the simplifications in (b), you should be able to quickly find the current values, which will allow you to calculate the voltages at the two ends of the charged capacitor. The voltage difference will give you the amount of charge on the fully-charged capacitor.

The charge on the capacitor a long time after the switch is closed is mC.

Added by Sophia K.

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