\begin{align*}P=IV\end{align*}
\begin{align*}R_{total} = R_1 + R_2 + R_3 + \ldots\end{align*}
\begin{align*}\frac{1} {R_{total}} = \frac{1} {R_1} + \frac{1}{R_2} + \frac{1} {R_3} + \ldots\end{align*}
Name  Electrical Symbol  Units  Analogy 

Voltage (\begin{align*}V\end{align*}  Volts \begin{align*}(V)\end{align*} 
A water dam with pipes coming out at different heights. The lower the pipe along the dam wall, the larger the water pressure, thus the higher the voltage. Examples: Battery, the plugs in your house, etc. 

Current (\begin{align*}I\end{align*}) 
Amps \begin{align*}(A)\end{align*} \begin{align*}A = \mathrm{C/s}\end{align*} 
A river of water. Objects connected in series are all on the same river, thus receive the same current. Objects connected in parallel make the main river branch into smaller rivers. These guys all have different currents. Examples: Whatever you plug into your wall sockets draws current 

Resistance (\begin{align*}R\end{align*})  Ohm \begin{align*}(\Omega)\end{align*} 
If current is analogous to a river, then resistance is the amount of rocks in the river. The bigger the resistance the less current that flows Examples: Light bulb, Toaster, etc. 
Example
A more complicated circuit is analyzed.
a. What is the total resistance of the circuit?
In order to find the total resistance we do it in steps (see pictures. First add the \begin{align*}90 \Omega\end{align*} and \begin{align*}10\Omega\end{align*} in series to make one equivalent resistance of \begin{align*}100\Omega\end{align*} (see diagram at below). Then add the \begin{align*}100\Omega\end{align*} to the \begin{align*}10\Omega\end{align*} in parallel to get one resistor of \begin{align*}9.1\Omega\end{align*}. Now we have two resistors in series, simply add them to get the total resistance of \begin{align*}29.1\Omega\end{align*}.
b. What is the total current coming out of the power supply?
Use Ohm’s Law \begin{align*}(V=IR)\end{align*} but solve for current \begin{align*}(I=V/R)\end{align*}. \begin{align*}I_{total}=\frac{V_{total}}{R_{total}}=20V/2.91\Omega=0.69\ Amps\end{align*}
c. What is the power dissipated by the power supply?
\begin{align*}P=IV\end{align*}, so the total power equals the total voltage multiplied by the total current. Thus, \begin{align*}P_{total}=I_{total}V_{total}=(0.69A)(20V)=13.8W\end{align*}.
d. How much power is the \begin{align*}20\Omega\end{align*} resistor dissipating?
The \begin{align*}20\Omega\end{align*} has the full 0.69Amps running through it because it is part of the ‘main river’ (this is not the case for the other resistors because the current splits). \begin{align*}P_{20 \Omega} = I^2_{20 \Omega} R_{20 \Omega} = (0.69A)^2 (20 \Omega) = 9.5W\end{align*}
e. If these resistors are light bulbs, order them from brightest to least bright.
The brightness of a light bulb is directly given by the power dissipated. So we could go through each resistor as we did the \begin{align*}20\Omega\end{align*} guy and calculate the power then simply order them. But, we can also think it out. For the guys in parallel the current splits with most of the current going through the \begin{align*}10\Omega\end{align*} path (less resistance) and less going through the \begin{align*}90\Omega+10\Omega\end{align*} path. Well the second path is ten times the resistance of the first, so it will have one tenth of the total current. Thus, there is approximately and 0.069 Amps going through the \begin{align*}90\Omega\end{align*} and \begin{align*}10\Omega\end{align*} path and 0.621Amps going through the \begin{align*}10\Omega\end{align*} path.
\begin{align*}P_{10 \Omega} &= I^2_{10\Omega}R_{10 \Omega}=(0.621A)^2(10\Omega)=3.8W\\ P_{90+10\Omega} &= I^2_{90+10\Omega}R_{90 + 10 \Omega}=(0.069A)^2(100\Omega)=0.5W\end{align*}
We now know that the \begin{align*}20\Omega\end{align*} is the brightest, \begin{align*}10\Omega\end{align*} is second and then the \begin{align*}90\Omega\end{align*} and last the \begin{align*}10\Omega\end{align*} (these last two have same current flowing through them, so \begin{align*}90\Omega\end{align*} is brighter due to its higher resistance).
\begin{align*}^*\end{align*}Note: Adding up these two plus the 9.5W from the \begin{align*}20\Omega\end{align*} resistor gives us 13.8W, which is the total power previously calculated, so we have confidence everything is good.
Interactive Simulations
Review
 What will the ammeter read for the circuit shown to the right?
 You can use the simulation below to check your answer. Click on the blue arrow and select the part of the circuit you want to track. Then scroll down to the Data tab and you can see the current and voltage in different parts of the circuit.
Draw the schematic of the following circuit.
 Analyze the circuit below.
 Find the current going out of the power supply
 How many Joules per second of energy is the power supply giving out?
 Find the current going through the \begin{align*}75\ \Omega\end{align*} light bulb.
 Find the current going through the \begin{align*}50\ \Omega\end{align*} light bulbs (hint: it’s the same, why?).
 Order the light bulbs in terms of brightness
 If they were all wired in parallel, order them in terms of brightness.
 Find the total current output by the power supply and the power dissipated by the \begin{align*}20\ \Omega\end{align*} resistor.
 You have a \begin{align*}600\;\mathrm{V}\end{align*} power source, two \begin{align*}10\ \Omega\end{align*} toasters that both run on \begin{align*}100\;\mathrm{V}\end{align*} and a \begin{align*}25\ \Omega\end{align*}resistor.
 Show me how you would wire them up so the toasters run properly.
 What is the power dissipated by the toasters?
 Where would you put the fuses to make sure the toasters don’t draw more than 15 Amps?
 Where would you put a \begin{align*}25\end{align*} Amp fuse to prevent a fire (if too much current flows through the wires they will heat up and possibly cause a fire)?
 Look at the following scheme of four identical light bulbs connected as shown. Answer the questions below giving a justification for your answer:
 Which of the four light bulbs is the brightest?
 Which light bulbs are the dimmest?
 Tell in the following cases which other light bulbs go out if:
 bulb \begin{align*}A\end{align*} goes out
 bulb \begin{align*}B\end{align*} goes out
 bulb \begin{align*}D\end{align*} goes out
 Tell in the following cases which other light bulbs get dimmer, and which get brighter if:
 bulb \begin{align*}B\end{align*} goes out
 bulb \begin{align*}D\end{align*} goes out
 Refer to the circuit diagram below and answer the following questions.
 What is the resistance between \begin{align*}A\end{align*} and \begin{align*}B\end{align*}?
 What is the resistance between \begin{align*}C\end{align*} and \begin{align*}B\end{align*}?
 What is the resistance between \begin{align*}D\end{align*} and \begin{align*}E\end{align*}?
 What is the the total equivalent resistance of the circuit?
 What is the current leaving the battery?
 What is the voltage drop across the \begin{align*}12\ \Omega\end{align*} resistor?
 What is the voltage drop between \begin{align*}D\end{align*} and \begin{align*}E\end{align*}?
 What is the voltage drop between \begin{align*}A\end{align*} and \begin{align*}B\end{align*}?
 What is the current through the \begin{align*}25\ \Omega\end{align*} resistor?
 What is the total energy dissipated in the \begin{align*}25\ \Omega\end{align*} if it is in use for 11 hours?
 You are given the following three devices and a power supply of exactly \begin{align*}120\;\mathrm{v}\end{align*}. \begin{align*}^*\end{align*} Device \begin{align*}X\end{align*} is rated at \begin{align*}60\;\mathrm{V}\end{align*} and \begin{align*} 0.5\;\mathrm{A}\end{align*}\begin{align*}^*\end{align*} Device \begin{align*}Y\end{align*} is rated at \begin{align*}15\;\mathrm{w}\end{align*} and \begin{align*}0.5\;\mathrm{A}\end{align*}\begin{align*}^*\end{align*} Device \begin{align*}Z\end{align*} is rated at \begin{align*}120\;\mathrm{V}\end{align*} and \begin{align*}1800\;\mathrm{w}\end{align*} Design a circuit that obeys the following rules: you may only use the power supply given, one sample of each device, and an extra, single resistor of any value (you choose). Also, each device must be run at their rated values.
Review (Answers)
 \begin{align*}0.5\mathrm{A}\end{align*}
 Answers will vary
 a. \begin{align*}0.94 \;\mathrm{A}\end{align*} b. \begin{align*}112 \;\mathrm{W}\end{align*} c. \begin{align*}0.35 \;\mathrm{A}\end{align*} d. \begin{align*}0.94 \;\mathrm{A}\end{align*} e. \begin{align*}50, 45, 75 \ \Omega\end{align*} f. both \begin{align*}50 \ \Omega\end{align*} resistors are brightest, then \begin{align*}45 \ \Omega\end{align*}, then \begin{align*}75 \ \Omega\end{align*}
 a. \begin{align*}0.76 \;\mathrm{A}\end{align*} b. \begin{align*}7.0 \;\mathrm{W}\end{align*}
 b. \begin{align*}1000 \;\mathrm{W}\end{align*}
 a. Bulb A b. Bulbs B and C c. i. B,C, and D go out ii. C goes out iii. A,B, and C stay on d. i. A will get dimmer and D will get brighter ii. A will get dimmer and both B and C will get brighter
 a. \begin{align*}9.1 \ \Omega\end{align*} b \begin{align*}29.1 \ \Omega\end{align*} c. \begin{align*}10.8 \ \Omega\end{align*} d.\begin{align*} 26.8 \ \Omega\end{align*} e. \begin{align*}1.8\mathrm{A}\end{align*} f. \begin{align*}21.5\mathrm{V}\end{align*} g. \begin{align*}19.4\mathrm{V}\end{align*} h. \begin{align*}6.1\mathrm{V}\end{align*} i. \begin{align*}0.24\mathrm{A}\end{align*} j. \begin{align*}16 \;\mathrm{kW}\end{align*}
 Answers will vary