Solve Electricity Equation: Physics Problem Help

In summary, a person is asking for help with solving a physics problem involving a circuit diagram and two capacitors. They are looking for the through and across variables, as well as the component equation for the single component with current i and voltage v. The solution involves finding the equivalent capacitance for the two capacitors in series, which is Ceq=C1C2/(C1+C2), and using Kirchoff's laws to determine the current and voltage equations. The correct answer is i=2C*dv/dt.
  • #1
Zurtex
Science Advisor
Homework Helper
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Hi, I've been asked by mother to help out one of her friends with a physics problem (she is doing a maths degree and got a physics question) however my knowledge of physics in this area appears to be even more limited. So if I write the problem out here could someone please point me in the right direction to give help.

I will write out everything I was given:
Diagram (electrical circuit)

Code:
 A     i[sub]1[/sub]                 B
  [b]·[/b]------->---| |---------[b]·[/b]
  |      --->V[sub]1[/sub]           |
  |                       |
  |  i[sub]2[/sub]                   |
  -------->---| |----------
      --->V[sub]2[/sub]

Want to represent the 2 capacitors as a single component,

What is the through variable? (current) [i.e i1 + i2]
What is the across variable? (voltage) [i.e v1 + v2]

What is the component equation for the single component with current i and voltage v?

(I thought this was [tex]i = c \frac{dv}{dt}[/tex] but it's not an option!)

Presumably this is a multiple-choice question, any help would be great thanks.
 
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  • #2
What is the through variable? (current) [i.e i1 + i2]
What is the across variable? (voltage) [i.e v1 + v2]

Looks like those are answered right there. :confused:

What is the component equation for the single component with current i and voltage v?

Perhaps they're looking for the equivalent capacitance? If so, then capacitors in series (as these are) are equivalent to a single capacitor with capacitance Ceq=C1C2/(C1+C2).
 
  • #3
Thanks I'll check it out with them :biggrin:
 
  • #4
Some more help requested

The answer to the capacitor question has answers 2i=Cdv/dt

or

i = 2C dv/dt

but I am unsure as to the correct choice.

It should be borne in mind that this question is not really about physics but is rather about modelling physical networks and is taught in general terms where the same formulation is used for various branches of science.

So we are using Kirchoff's laws in general terms, given three sets of equations, component equations, vertex law equations and the cycle law equations.

Tired. :zzz:
 
Last edited:
  • #5
TiredTeddy said:
The answer to the capacitor question has answers 2i=Cdv/dt

or

i = 2C dv/dt

but I am unsure as to the correct choice.

Well, I am not sure about the terms used in the problem, but usually A, B, ... mean nodes in a electric circuit. You can imagine that the circuit in the figure is a part of a bigger net. The two capacitors are connected in parallel with respect to A and B. For parallel connected components, the voltages are the same. The voltage is the same across both capacitors V(AB)=V1=V2 and the current which flows through the single component between A and B is the sum of i1 and i2, i(AB) = i1 + i2. The component equation for a capacitor is Q=C*V, and i = dQ/dt=C*dV/dt--->i= i1+i2 = C1*dV/dt + C2*dV/dt = (C1+C2)*dV/dt = C(AB)*dV/dt for the single component capacitor between A and B. If both capacitances are equal i = 2C*dv/dt.

ehild
 

1. How do I solve an electricity equation?

To solve an electricity equation, you need to apply the basic principles of electricity, such as Ohm's Law and Kirchhoff's Laws. You also need to have a good understanding of the variables involved, such as voltage, current, and resistance. Once you have identified the variables and their relationships, you can use algebraic equations to solve for the unknown values.

2. What is Ohm's Law and how is it used in solving electricity equations?

Ohm's Law states that the current through a conductor is directly proportional to the voltage and inversely proportional to the resistance. This means that as voltage increases, current increases, and as resistance increases, current decreases. Ohm's Law is used to determine the relationship between these variables and can be rearranged into different equations to solve for the unknown values.

3. How do I apply Kirchhoff's Laws in solving electricity equations?

Kirchhoff's Laws are two principles that are used to analyze electrical circuits. The first law states that the sum of currents entering a junction must equal the sum of the currents leaving the junction. The second law states that the sum of voltage drops in a closed loop must equal the sum of voltage rises. These laws can be used to set up equations and solve for unknown variables in a circuit.

4. What are the common mistakes to avoid when solving electricity equations?

One common mistake is to forget to convert units when necessary. For example, if the resistance is given in kilohms, but the equation requires resistance in ohms, the value must be converted before solving. Another mistake is to use the wrong formula or equation for a given situation. It's important to carefully read the problem and understand which principles and equations are applicable.

5. Are there any tips or tricks for solving electricity equations more efficiently?

One tip is to draw a diagram or circuit representation of the problem to help visualize the relationships between the variables. Another tip is to start with the simplest equations and work your way up to more complex ones. It's also helpful to double-check your calculations and units to avoid mistakes. Finally, practice and familiarity with the concepts and equations will ultimately lead to more efficient problem-solving.

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