Question about Capacitance and Inductance

In summary, the conversation revolves around two problems related to capacitance and inductance. The first problem involves converting the number of calories in a burger to joules and determining the voltage on a 1.1 Farad capacitor. The second problem deals with finding the current in milliamps when all the energy from a charged capacitor is transferred to an inductor. The equations used are E = 0.5CV^2 and I = √(2E/L). There were some errors in the calculations, particularly in converting calories to joules, but the final answer for the second problem was 362.766 amps.
  • #1
Mohamed Abdul

Homework Statement


[/B]
Hey everyone, I recently got two problems on capacitance and inductance and need some help on them. Here are the two problems:

1. A Royal Red Robin Burger contains 1230 food calories. Also, more that the daily requirement for protein, fat, sodium ..., even without fries or a drink. Convert this number of calories to Joules. For the amount of energy you have calculated, determine the voltage V (in kVolts) on a capacitor C = 1.1 Farads.

2. A capacitor C = 33 micro-Farads is charged up to a voltage V = 16.1 Volts. It is then connected to an inductor L = 65 milli-Henries using a switch. The energy will oscillate back and forth between the capacitor and inductor. When all of the energy is in the inductor, what current I will it be carrying? Express your answer in milli-Amps.

Homework Equations


E = .5CV^2 I=sqrt(2E/L)

The Attempt at a Solution



For the first question, I multiplied 1230 by 4.18, and set that number (5141.4) equal to .5(1.1)(V^2). Then I solved for V and got /0967 kVolts.

For the second one I multiplied .5*.000033 farads*16.1^2 and got .03888 Joules. Then I set that equal to the second equation and substituted E and L and got 345.88 milliamps.

I want to know if I'm on the right track for this problem, since this is my final allowed attempt at it.
 
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  • #2
Mohamed Abdul said:
For the first question, I multiplied 1230 by 418
418 ??
For the second one I multiplied .5*.000033 farads*16.1^2 and got .03888 Joules.
I got 4.28e-3 joules.
 
  • #3
rude man said:
418 ?? Better check that second equation ...
Is that not the equation for current. If so, then what is? Also I got the 418 because there are 4.18 joules per calorie and I forgot the decimal, sorry,
 
  • #4
rude man said:
I got 4.28e-3 joules.
I redid my math and got that as well, my final answer for question 2 is now 362.766 amps. Did that line up with what you got?
 
  • #5
Mohamed Abdul said:
Is that not the equation for current. If so, then what is? Also I got the 418 because there are 4.18 joules per calorie and I forgot the decimal, sorry,
I thought I edited that cooment out. Your eq. is correct, the math was not.
 
  • #6
Mohamed Abdul said:
I redid my math and got that as well, my final answer for question 2 is now 362.766 amps. Did that line up with what you got?
Exactly.
 
  • #7
I'm still wondering if I did the right steps on the first problem, as I've constantly been making errors on those types of questions. I'm not sure if its due to a rounding mistake or a genuine error.
 
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  • #8
Mohamed Abdul said:
I'm still wondering if I did the right steps on the first problem, as I've constantly been making errors on those types of questions. I'm not sure if its due to a rounding mistake or a genuine error.
You still did not do the calories to joules conversion correctly.
Fact: 1 food calorie = 1000 cgs calories. 1 food calorie = 4.18 kilojoules, not joules.
 

FAQ: Question about Capacitance and Inductance

What is the difference between capacitance and inductance?

Capacitance and inductance are both electrical properties of a circuit, but they have opposite effects. Capacitance is the ability of a component to store electrical charge, while inductance is the ability to resist changes in current flow. In other words, capacitance stores energy, while inductance resists the flow of energy.

How are capacitance and inductance measured?

Capacitance is measured in units of farads (F) and is usually denoted by the letter C. It can be measured using a capacitance meter or by using a known voltage and measuring the resulting charge. Inductance is measured in units of henries (H) and is denoted by the letter L. It can be measured using an inductance meter or by using a known current and measuring the resulting magnetic field.

What is the relationship between capacitance and inductance in a series circuit?

In a series circuit, capacitance and inductance have a reciprocal relationship. This means that as capacitance increases, inductance decreases and vice versa. This relationship can be represented by the equation XL = 1/(2πfC), where XL is the inductive reactance, f is the frequency, and C is the capacitance.

How do capacitance and inductance affect the behavior of AC circuits?

Capacitance and inductance play important roles in the behavior of AC circuits. Capacitors and inductors can be used to manipulate the phase and amplitude of AC signals, and they can also affect the frequency response of a circuit. In AC circuits, capacitance and inductance create a phenomenon known as reactance, which is the opposition to the flow of AC current.

How are capacitance and inductance used in electronic devices?

Capacitors and inductors are commonly used in electronic devices for various purposes. For example, capacitors are used to filter out unwanted noise in power supplies, while inductors are used in power converters to regulate the output voltage. They are also used in timing circuits, oscillators, and filters for signal processing. In addition, the combination of capacitance and inductance is used in resonant circuits, such as radio frequency (RF) filters and antennas.

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