Complete+Solution+Chart+for+Circuits

=Complete Solution Charts for Circuits=

//from dvapphysics.wikispaces, the free AP Physics wiki that anyone in Mr. Adkin's class can edit.//

To give a general understanding what a circuit is: a electronic circuit is composed of individual electronic components, such as resistors and capacitors, connected by conductive wires or traces through which electronic current can flow.

The **Complete Solution Chart** (CSC) is a comprehensive method of solving problems involving circuit diagrams by organizing the various aspects of the electronic circuit. This method was created by Mr. Adkins and isn't very familiar outside the classroom. To some individuals, the chart can show that you're just being neat and tidy about showing your work.

toc =Resistors and Capacitors=

Capacitors
The **capacitance** C of a capacitors is the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between conductors:
 * = [[image:MSP10519fii6cecaec3aia00005e8f7gebgagc9eaa.gif width="71" height="48"]] ||
 * = C ||= Capacitance ||
 * = Q ||= Charge ||
 * = V ||= Voltage ||


 * ~  ||= C ||= Q ||= V ||= E ||
 * ~ Series ||= 1/C ||= = || + || + ||
 * ~ Parallel ||= + ||= + ||= = || + ||

Resistors
The **resistance** (Ω) R of a conductor is defined as the ratio of the potential difference across the conductor to the current in it. It's basically the amount of resistance an object resists or opposes the flow of current:

The triangle above is a handy way of remembering three equations: //Voltage equals current times resistance.// //Current equals voltage over resistance.// Resistance equals voltage over current.
 * [[image:0098.png width="77" height="19"]] ||
 * V || voltage ||
 * I || electrical current ||
 * R || electrical resistance ||
 * V = IR
 * I = V/R
 * R= V/I
 * ~  ||= V ||= I ||= R ||= P ||
 * ~ Series ||= + ||= = || + || + ||
 * ~ Parallel ||= = || + || 1/R || + ||

=Series, Parallel, and Combination= //Examples of how to use charts for series, parallel and a combination of the two .//

Series
In a series circuit, as you can see from the picture below, shows that it is an electric circuit connected so that current can flow without having to branch off. From the picture shown above, the voltage is 10V and R1 and R2's unit are Ohms which can be assumed to be resistors. For this picture, we will be using the chart for V = I R for series. First, we can fill in what we already know. We know that the total voltage is 10 and the total electrical resistance will be 6 Ω because resistance adds up all resistors in a series circuit. Next, to find the total electrical current flowing through the circuit, you have to divide the total voltage by the total electrical resistance, I = V/R.
 * ~  ||= V ||= I ||= R ||= P ||
 * ~ Series ||= + ||= = || + || + ||
 * ~  || V || I || R || P ||
 * ~ R1 ||  ||   || 5Ω ||   ||
 * ~ R2 ||  ||   || 1Ω ||   ||
 * ~ RT || 10V ||  || 6Ω ||   ||

I = V/R I = 10/6 ~ 5/3

Since in the column for Current(I) has a (=) sign, you're answer for total electrical current will be the same in the cells for R1 and R2. Next, to find the amount of voltage that is going through the first and second resistor separately, you would have to multiply the electrical current and resistance for each row, V = IR.
 * ~  || V || I || R || P ||
 * ~ R1 ||  || 5/3A || 5Ω ||   ||
 * ~ R2 ||  || 5/3A || 1Ω ||   ||
 * ~ RT || 10V || 5/3A || 6Ω ||  ||

R1: V = IR V = (5/3)(5) V = 25/3

R2: V = IR V = (5/3)(1) V = 5/3 Finally, to find P, the power, you will have to multiply the voltage and electrical current for each row, P= IV.
 * ~  || V || I || R || P ||
 * ~ R1 || 25/3V || 5/3A || 5Ω ||  ||
 * ~ R2 || 5/3V || 5/3A || 1Ω ||  ||
 * ~ RT || 10V || 5/3A || 6Ω ||  ||

R1: P = IV P = (5/3)(25/3) P = 125/9

R2: P = IV P = (5/3)(5/3) P = 25/9

RT: P = IV P = (10)(5/3) P = 50/9 When you added up the last column of Power and multiply the total amounts for voltage and current, you should reach the same answer and in this case, 50/3 was the correct answer.
 * ~  || V || I || R || P ||
 * ~ R1 || 25/3V || 5/3A || 5Ω || 125/9 ||
 * ~ R2 || 5/3V || 5/3A || 1Ω || 25/9 ||
 * ~ RT || 10V || 5/3A || 6Ω || 150/9 ~ 50/3 ||

Parallel
In a parallel circuit, as you can see from the picture below, shows that it is an electric circuit connected so that current can flow and branch off in one or more directions while the current is going one way. From the picture shown above, the voltage is 18V and R2 is branched off from the circuit. For this picture, we will be using the the chart for parallel. First, we can fill in what we already know. We know that the total voltage is 18V and the total electrical resistance will be calculated by a side equation.
 * ~  ||= V ||= I ||= R ||= P ||
 * ~ Parallel ||= = || + || 1/R || + ||

1/RT = 1/R1 + 1/R2 1/RT = 1/6 + 1/3 1/RT = 1/6 + 2/6 1/RT = 3/6 RT = 6/3 RT = 2Ω

Since in the column for Voltage(V) has a (=) sign, you're answer for total voltage will be the same in the cells for R1 and R2. Next, to find the the total electrical current, you will have to use the equation I = V/R. Finally, to find the Power, use the equation P = IV.
 * ~  || V || I || R || P ||
 * ~ R1 || 18V ||  || 6Ω ||   ||
 * ~ R2 || 18V ||  || 3Ω ||   ||
 * ~ RT || 18V ||  || 2Ω ||   ||
 * ~  || V || I || R || P ||
 * ~ R1 || 18V || 3A || 6Ω ||  ||
 * ~ R2 || 18V || 6A || 3Ω ||  ||
 * ~ RT || 18V || 9A || 2Ω ||  ||
 * ~  || V || I || R || P ||
 * ~ R1 || 18V || 3A || 6Ω || 54 ||
 * ~ R2 || 18V || 6A || 3Ω || 108 ||
 * ~ RT || 18V || 9A || 2Ω || 162 ||

Combination
In a combination circuit, as you can see from the picture below, is a combination of a series and a parallel circuit. For this example, we will be required to use both the series and parallel charts. //Note: Do not blend resistors. As in:// //It may get the same answer in the very end for you but on tests or questions, it doesn't always ask for ALL// //everything that is in the chart. It may ask for one or two, maybe three specific parts and you want to make sure// //that you know how to solve for each component so you don't have to waste your time doing too much work for// //only one answer.//
 * ~ . ||= V ||= I ||= R ||= P ||
 * ~ Series ||= + ||= = || + || + ||
 * ~ Parallel ||= = || + || 1/R || + ||
 * . ||= //V// || //I// || //R// || //P// ||
 * //R1 + R3// |||||||| <--- //do not do this.// ||
 * //R2// ||  ||   ||   ||   ||
 * RT ||  ||   ||   ||   ||

Now to solve the circuit, you will have to do two parts. The first part is to choose the section to work on first. The section that would have to be done first is the branched off parts of the circuit, which will be solved by the parallel circuit chart. From the picture, we can fill in what's given to us. Then we have to solve for how much resistance using the 1/RT = 1/R1 = 1/R2 equation.
 * ~  ||= V ||= I ||= R ||= P ||
 * ~ Parallel ||= = || + || 1/R || + ||
 * ~ . ||= //V// || //I// || //R// || P ||
 * ~ //R1// ||  ||   || 5Ω ||   ||
 * ~ //R3// ||  ||   || 5Ω ||   ||
 * ~ RT || 20V ||  ||   ||   ||

1/RT = 1/R1 + 1/R2 1/RT = 1/5 + 1/5 1/RT = 2/5 RT = 5/2Ω

Since in the column for Voltage(V) has a (=) sign, you're answer for total voltage will be the same in the cells for R1 and R3. Next, we'll use the equation I =V/R to solve how much electrical current is going through each resistor. Moving on from parallel to series: Using what we solved from the parallel circuit chart, we can plug that in with what we know in the series circuit chart. After plugging in what we know, we'll use the equation I = V/R To solve for V, we use the equation V = IR. Finally, we can solve for power with P = IV
 * ~ . ||= //V// || //I// || //R// || P ||
 * ~ //R1// || 20V ||  || 5Ω ||   ||
 * ~ //R3// || 20V ||  || 5Ω ||   ||
 * ~ RT || 20V ||  || 5/2Ω ||   ||
 * ~ . ||= //V// || //I// || //R// || P ||
 * ~ //R1// || 20V || 4A || 5Ω || 80 ||
 * ~ //R3// || 20V || 4A || 5Ω || 80 ||
 * ~ RT || 20V || 8A || 5/2Ω || 160 ||
 * ~  ||= V ||= I ||= R ||= P ||
 * ~ Series ||= + ||= = || + || + ||
 * ~ . ||= //V// || //I// || //R// || P ||
 * ~ //R2// ||  ||   || 10Ω ||   ||
 * ~ //R4// ||  ||   || 5/2Ω ||   ||
 * ~ RT || 20V ||  || 10 5/2Ω ||   ||
 * ~ . ||= //V// || //I// || //R// || P ||
 * ~ //R2// ||  || 1.6A || 10Ω ||   ||
 * ~ //R4// ||  || 1.6A || 5/2Ω ||   ||
 * ~ RT || 20V || 1.6A || 10 5/2Ω ||  ||
 * ~ . ||= //V// || //I// || //R// || P ||
 * ~ //R2// || 4 || 1.6A || 10Ω ||  ||
 * ~ //R4// || 16 || 1.6A || 5/2Ω ||  ||
 * ~ RT || 20V || 1.6A || 10 5/2Ω ||  ||
 * ~ . ||= //V// || //I// || //R// || P ||
 * ~ //R2// || 4 || 1.6A || 10Ω || 6.4 ||
 * ~ //R4// || 16 || 1.6A || 5/2Ω || 25.6 ||
 * ~ RT || 20V || 1.6A || 10 5/2Ω || 32 ||

//Reference:// //College Physics Textbook//