Power loss in wires
P(loss in wire) = [I^2 * Voltage drop at the wire] (Not Vterminal)
Voltage difference for capacitors
V= E(lectric Field) * d V= Ed
E stored in capacitor
E = 1/2 Q V
U = (1/2) CV^{2}
C of capacitor (capacitance)
C= k(Ę)A/d (Let Ę mean epsilon naught constant)
Dialectic Constant (k)
Medium between capacitor plates *all about polarizability *more polar the medium, the higher the k Vacuum k=1 Glass k=5 Water k=80 (HUGE)
How do you raise Q for a given battery? (3 ways, hint: capacitance)
1) decrease distance between plates 2) increase area of plates 3) raise dielectric constant
Capacitors & time (graph: I = y, x = t)
Slope: Current will decrease exponentially with time Area: I x t= C/s x s = C
Resistors wired in series
Current through each R is equal. Voltage drop on each at is a fraction of the terminal voltage. Voltage drop on each R is PROPORTIONAL to resistance I(1) = I(2) V(1) = I(1)*R(1) V(2) = I(2) * R(2) R=R(1) + R(2) + R(3)
Capacitors wired in series

"SeriQ"; V_{eq} is the sum of all individual V's
 Q_{1}=Q_{2}
 V_{s}=V_{1}+V_{2}+V_{3}+...+V_{n}
V is proportional to 1/C
 V_{1}=Q_{1}/C_{1}
 V_{2}=Q_{2}/C_{2}
 1/C_{eq} = (1/C_{1}) + (1/C_{2})+...+(1/C_{n})
Capacitors wired in parallel
 "Par=C" ?
 resultant capacitance that is equal to the sum of the individual capacitances
 C_{p}=C_{1} + C_{2 }+ C_{3}+...+C_{n}
 V_{p}=V_{1}=V_{2}=V_{3}=...V_{n}
V of capacitor
Q/C
V of resistor (Vr)
Vr=IR
V terminal of RC circuit
V terminal = V of capacitor + V of resistor Vt=Vr + Vc
conventional current
convention assumes that all moving charges are positive, and thus, conventional current moves from regions of high electric potential to regions of low electric potential.
In the figure above, the conventional current moves through the circuit, from the positive terminal to the negative terminal.
direct current
where Δ q is the amount of charge that passes by in a certain amount of time, Δ t.
magnitude stays constant over time
electromotive force (EMF)
the electric potential difference (or voltage) that drives current. EMF isn't really a force, it's a potential difference.
Kirchhoff's Loop Rule
The sum of the voltage drops across a closed circuit has to equal the sum of the voltage gains across the circuit.
resistance
of a conductor, R, is a measure of a particular conductor's resistance to current flow
resistors.
Devices used to intentionally cause a voltage drop
resistivity, ρ
of a material is a measure of its resistance to current flow; if a material has a higher resistivity, then applying a given potential difference across it will produce a lower current. The resistance of a resistor depends of the resistivity of the material:
where A is the crosssectional area through which the current flows, and L is the length of the resistor
ohmic
the voltage drop across them will be proportional to the current passing through them (conditions on MCAT unless otherwise stated)
Ohm's Law
V=IR
Electric current
ampere
current units are coulombs per unit time
From Ohm's Law, we see that the units of resistance must then be a volt per unit ampere, or an ohm
Resistors in Series configuration
When two or more resistors are in a series configuration, the current has no choice but to pass through all of them, one after the other.
In the diagram above, resistors R1 and R2 are in series. If you replaced the two series resistors by a single equivalent resistor, the total resistance would be: R_{eq} = R_{1} + R_{2}
 What is the total resistance of the above circuit in ohms?
 How much current passes through each of the first and second resistors?
 How much current passes through the whole resistor?
 4 ohms
 2.5 A and 2.5 A
 2.5 A
The equivalent resistance of the two series resistors is:
R_{eq} = 1 Ω + 3 Ω = 4 Ω
Now we can calculate the total amount of current flowing out of and into the battery, using Ohm's Law:
Because the two resistors are in series, all 2.5 A of current passes through each of the resistors as it passes through the circuit.
The voltage drops across each resistor will be different. Across the 1Ω resistor, the voltage drop is V1 = IR = (2.5 A)(1Ω ) = 2.5 V. Across the 3Ω resistor, the voltage drop is V3 = IR = (2.5 A)(3Ω ) = 7.5 V. Notice that Kirchhoff's loop rule still applies: the total voltage gain, due to the battery, is 10 V, and the total voltage drop across the two resistors is the same: 2.5 V + 7.5 V = 10 V.
Resistors in Parallel
When two or more resistors are in a parallel configuration, the current is split between the different paths, the voltage across each path is the same.
Req = (R_{1}R_{2}) / (R_{1}+R_{2})
 What is the total resistance of the above circuit in ohms?
 How much current passes through the whole circuit?
 How much current passes through the 2Ω resistor?
 How much current passes through the 5Ω resistor?
 10/7 Ohms
 2.8 A
 2 A
 0.8 A
The equivalent resistance of the two parallel resistors can be calculated:
Now we can calculate the total amount of current flowing out of and into the battery, using Ohm's Law:
Because the two resistors are in parallel, part of the 2.8 A of current passes through the 2Ω resistor, and the other part passes through the 5Ω resistor.
However, the voltage drops across each resistor are the same. Across the 2Ω resistor, the current is:
Since the total current through the circuit is 2.8A, and 2A pass through the first circuit, we can use Kirchhoff's junction rule to calculate the current through the second circuit. Knowing that the total current going through the circuit is equal to the sum of the currents going through each resitor, we can calculate the current through the second resitor to be 2.8A  2.0A = 0.8A.
What is the equivalent resistance seen by the 25V battery, in Ohms?
250 Ohms
The 160Ω and the 200Ω resistors at the top of the circuit are in series: . This is parallel with the 120Ω resistor:
So the total equivalent resistance of the circuit is:
What current comes out of the battery?
0.1 A
Since the equivalent resistance of the circuit is 250 Ω , the total current in the circuit is: