Electric Circuit: Case I & II Solutions

In summary: However, it would be helpful to show your work for part d and e to ensure accuracy. In summary, the circuit includes a switch S that can connect or disconnect a 3-microfarad capacitor in parallel with a 10-ohm resistor. In Case I, where the switch is open, the current in the battery is 8 amps, the current in the 10-ohm resistor is 2 amps, and the potential difference across the 10-ohm resistor is 20 volts. In Case II, where the switch is closed and the currents reach constant values, the charge on the capacitor is 6x10-5 Coulombs and the energy stored in the capacitor is 6x10-4 Joules
  • #1
livewire1717
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0

Homework Statement


http://img88.imageshack.us/my.php?image=physicsxt0.jpg

The circuit shown above includes a switch S, which can be closed to connect the 3-microfarad capacitor in parallel with the 10-ohm resistor or opened to disconnect the capacitor from the circuit.

Case I. Switch S is open. The capacitor is not connected. Under these conditions determine:
a) The current in the battery
b) the current in the 10-ohm resistor
c) the potential difference across the 10-ohm resistor

Case II. Switch S is closed. The capacitor is connected. After some time, the currents reach constant values. Under these conditions determine:
d) The charge on the capacitor
e) The energy stored in the capacitor

The Attempt at a Solution


part a)I found the total resistance to be 9 ohms: 6+ 1/(1/4+1/(10+2)) so I=72/9=8 amps
Part b) I thought that 1/4 of the current would go through the 10 ohm resistor (12 ohms total in that branch, 4 in the parallel branch) so 1/4 of 8 is 2 amps
Part c) V=IR, V=(2 amps)(10 ohms)=20 V
Part d) The voltage drop should be the same as in the 10 ohm resistor, Q=(3x10-6)(20)=6x10-5C
Part e) U=QV/2, U=(6x10-5)(20)/2=6x10-4J

I think part b might be wrong, which screwed up the rest.. Can anyone either confirm this or correct me? Thanks!
 
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  • #2
Your answer for part b is correct. I used Kirchoff's voltage law and current nodes. It appears the rest of your calculations are correct also.
 
  • #3


I would like to provide a response to your solution attempt.

Firstly, your calculation for part b is incorrect. The current through the 10-ohm resistor is not 1/4 of the total current, but rather the same as the total current (8 amps). This is because the capacitor is not connected in this case, so it does not affect the current flow in the circuit. Therefore, the current in the 10-ohm resistor is also 8 amps.

Secondly, your calculation for part c is correct. The potential difference across the 10-ohm resistor is indeed 20V.

Moving on to Case II, your calculation for part d is also correct. The charge on the capacitor is 6x10^-5C.

However, your calculation for part e is incorrect. The energy stored in a capacitor is given by the equation U = 1/2CV^2, where C is the capacitance and V is the potential difference across the capacitor. In this case, the potential difference is 20V, but the capacitance is 3 microfarads (3x10^-6F). Therefore, the energy stored in the capacitor is U = (1/2)(3x10^-6)(20)^2 = 6x10^-4J.

Overall, your solution attempt was mostly correct except for the mistake in part b and the incorrect equation used in part e. It is important to carefully consider the effects of the switch and the components in the circuit when solving for different cases. Great job on attempting the problem, and keep up the good work!
 

Related to Electric Circuit: Case I & II Solutions

1. How do you calculate voltage in an electric circuit?

Voltage can be calculated by multiplying the current (in amperes) by the resistance (in ohms) in a circuit. This is known as Ohm's Law: V=IR. It is important to note that voltage is measured in volts and can also be written as V=E/Q, where E represents energy and Q represents charge.

2. What is the purpose of a resistor in an electric circuit?

A resistor is a passive component in a circuit that is used to resist the flow of current. It is often used to control the amount of current and voltage in a circuit. Resistors can also be used to limit the amount of current flowing through a specific component to prevent damage.

3. How do you determine the total resistance in a series circuit?

In a series circuit, the total resistance is equal to the sum of all individual resistances. This can be calculated by using the formula Rtotal = R1 + R2 + R3...+ Rn, where R represents the resistance of each component in the circuit.

4. What is the difference between a parallel circuit and a series circuit?

In a parallel circuit, the components are connected in branches, allowing multiple paths for the current to flow. This results in a lower overall resistance and higher total current compared to a series circuit, where the components are connected in a single loop.

5. How do you troubleshoot a circuit that is not working?

First, check for any loose or damaged connections and make sure the circuit is properly grounded. Use a multimeter to check for continuity and ensure that all components are functioning properly. If necessary, replace any damaged components. It may also be helpful to refer to the circuit diagram to ensure all components are connected correctly.

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