Alternating current circuits

In summary, the conversation discusses a light bulb with a resistance of 212 connected to a wall socket with 120 V and 60.0 Hz. The question is to determine the current in the bulb and after adding a 13.2-μF capacitor in series. The equations used are X(sub L)=2pifL and Irms=Vrms/Z, with Z being the impedance of the circuit. The question is raised about the necessity of another equation, but it is clarified that Z is the impedance.
  • #1
25
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Homework Statement


A light bulb has a resistance of 212 . It is connected to a standard wall socket (120 V, 60.0 Hz). (a) Determine the current in the bulb. (b) Determine the current in the bulb after a 13.2-μF capacitor is added in series in the circuit.

Homework Equations


R, V, f are given. I don't know how to start incorporate these figures without a value for Z. Am I forgetting a another important equation?


The Attempt at a Solution


I used the equations:
X(sub L)=2pifL
Irms=Vrms/Z
 
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  • #2
arod2812 said:

The Attempt at a Solution


I used the equations:
X(sub L)=2pifL

Why? There is no inductor in the circuit!
arod2812 said:
Irms=Vrms/Z

Okay, good. Z is the impedance. What is the impedance of the circuit with just the resistor? What about after adding a capacitor?
 
  • #3

Z=sqrt(R^2 + X(sub L)^2)

(a) To determine the current in the bulb without the capacitor, we can use the equation Irms=Vrms/Z, where Vrms is the root mean square value of the voltage (120 V) and Z is the total impedance of the circuit. Since the bulb has a resistance of 212 Ω, we can calculate the impedance as Z=sqrt(R^2 + X(sub L)^2), where X(sub L) represents the inductive reactance of the circuit, given by X(sub L)=2pifL. The frequency is given as 60.0 Hz and the inductance of the circuit can be assumed to be negligible, so X(sub L)=0. Therefore, the total impedance is simply equal to the resistance, Z=212 Ω. Plugging in these values, we get Irms=120 V/212 Ω=0.566 A. Therefore, the current in the bulb without the capacitor is 0.566 A.

(b) To determine the current in the bulb after the capacitor is added in series, we can use the same equation, Irms=Vrms/Z, but this time the impedance will be different. Since the capacitor is now in series, the total impedance of the circuit will be the sum of the resistance and the capacitive reactance, given by X(sub C)=1/(2pifC). The frequency is still 60.0 Hz, but now we have a capacitance of 13.2 μF, which can be converted to Farads as 13.2x10^-6 F. Plugging in these values, we get X(sub C)=1/(2pifC)=1/(2x3.14x60.0x13.2x10^-6)=201.5 Ω. Therefore, the total impedance is now Z=212 Ω + 201.5 Ω=413.5 Ω. Plugging this into the equation, we get Irms=120 V/413.5 Ω=0.290 A. Therefore, the current in the bulb with the capacitor added is 0.290 A. We can see that the current has decreased due to the addition of the capacitor, which acts as a reactive element in the circuit and decreases the overall impedance.
 

1. What is alternating current (AC) and how does it differ from direct current (DC)?

Alternating current (AC) is a type of electrical current where the direction of flow of electric charge changes periodically. This is in contrast to direct current (DC), where the flow of electric charge is in one direction only.

2. How does an alternating current circuit work?

An alternating current circuit works by using a source of AC electricity, such as a generator or power grid, to supply a constantly changing voltage to a load. This voltage alternates between positive and negative, causing the flow of electric charge to also alternate in direction.

3. What are the components of an alternating current circuit?

The main components of an alternating current circuit include a source of AC electricity, such as a generator or power grid, a load, which is the device that uses the electricity, and various electrical components such as resistors, capacitors, and inductors that control the flow of electricity.

4. What are the advantages of using alternating current circuits?

One of the main advantages of alternating current circuits is that they are able to efficiently transmit electricity over long distances. AC voltage can be easily increased or decreased using transformers, making it more suitable for long-distance power transmission compared to DC circuits. Additionally, AC allows for the use of different types of electrical components, making it more versatile for a variety of applications.

5. How are alternating current circuits used in everyday life?

Alternating current circuits are used in a wide range of everyday devices, including household appliances, electronic devices, and power tools. They are also used in lighting systems, electric motors, and power grids. In short, any device that requires electricity to function likely uses an alternating current circuit.

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