General Electricity Notes: • DC current--current flows in one direction o Current is defined as the rate of flow of charge o Units: Amp(A), 1A=1C/s o Conventional current-- flows form positive to negative terminal o Electron flow--negative to positive terminal. Electrons are the mobile charge carriers in a conductor o The charge of an electron is -1.6x10-19C o Formulas: § I=Q/t, I=current(A), Q= charge(C), t=time(s) • Voltage o It is also known as potential difference o The electrical potential difference is defined as the amount of electric potential energy (equivalent to work), that is gained or lost when a unit electric charge is moved from one point to another. o Units:Volt (V), 1V=1 J/C § W=E=VQ, W=work(J), E=energy(J),V= voltage(V), Q=charge(C) • Resistance o Electrical resistance is an indicator of the degree to which electrical current is restricted. o The resistance of an object depends on the material and the geometry (size and shape). A thinner wire will have more resistance than a thicker made of the same material. A longer wire would have more resistance than a shorter one. o Materials which are better conductors, allow the electrons to move more freely. Materials which are better insulators, impede the flow of electrons. o Equivalent (or total) resistance: A group of resistors can be replaced by a single equivalent resistance, which would draw the same current as the group with the same applied voltage. o Units: Ohm(Ω) • Ohms Law o The Voltage across a given resistor of resistance R is proportional to the current through it. o Formula: § V=IR, V=Voltage(V), I=current(A), R=resistance (Ω) • Resisters in Series o There is one path for the current to follow, so the current is the same through each resistor o Formula: § Rt=R1+R2+R3..., Rt= equivalent resitance(Ω) • Resistors in parallel o Because of Kirchhoff''s voltage law, resistors in parallel have the same voltage across them o Formula § 1/Rt=1/R1+1/R2+1/R3 • Power and Energy o The electrical Power and energy used by a resistor o Formulas: § P=VI=V2/R=I2R, P=Power (W), V=voltage(V), I=current(A), R=resistance(Ω) § E=Pt, E=Energy (J), t=time(s) • Kirchhoff Laws o Voltage Law: The sum of all the voltage drops and increases around a complete path is equal to zero. In other words, around a compete path, the voltage supplied by the battery is equal to the sum of all the voltage drops across a resistor o Current Law (Junction Rule): The total current which enters a junction must equal the total current which leaves the junction Other notes: • Voltmeter-- A voltmeter is use to measure the voltage between two points in a circuit. A voltmeter is connected in parallel not in series; • Ammeter-- An ammeter measures the current which flows through a given path in the circuit. It should be connected in series, not parallel: • Adding resistors in parallel will reduce to the overall resistance of a circuit where as adding them in series will tend to increase the overall resistance. • In the circuit (right), if one were to keep adding resistors in parallel (at the dotted lines) the total resistance would decrease but could never decrease below the value of R. As the additional resistors are added the resistance of the group RG in parallel would approach zero. The total resistance of the circuit Rt= R+RG, ∴Rt→R, as more resistors are added General Combination Circuit Problem diagram 1 a) Determine the total resistance of the circuit b) What current leaves the battery? c) Determine the voltage across, and the current through every resistor. d) What was the energy used by the entire circuit, and by the 2Ω resistor in a time of 6 hours? Solution: a) • Label conventional currents (+ terminal to - through circuit) diagram 2 • Reduce circuit in steps. Draw equivalent circuits. The 1Ω and 2Ω resistors are clearly in series (R= 1+2 = 3Ω) The 3Ω and 6Ω are clearly in parallel (1/R=1/6+1/3, R= 2Ω) It is now clear that the all the remaining resistors are in series, giving us our final equivalent circuit with the total equivalent resistance. Rt= 11Ω b) The current which leaves the battery, using Ohm's law is: I=V/R= 88/11= 8A c) • Referring to circuit diagram 2, one can see now that we know the current through the 4Ω and 5Ω resistors: 8A • Using Ohms law (V=IR) one can calculate the voltages across these resistors. They are 32V and 40V respectively. • Now using Kirchoff's voltage law one can state that: o 88=32+V6+40, where V6 is the voltage across the 6Ω resistor o Therefore, V6= 16V • Now knowing the voltage V6 one can determine the current I2through the 6Ω resistor. o I2= V/R=16/6=2.67A, Now using the other Kirchoff Law (Junction rule) one can find I1 o It= I1 + I2 , 8= I1 + 2.6667 ∴ Ι = 5.33 A, remember to carry extra sig figs with intermediate calculation! o 1 I1 is the current through both the 1Ω and 2Ω resistors (in series) Using Ohm's Law one can calculate the voltages across these resistors (V=IR) o The voltages are 5.33V and 10.7V respectively. • test d) Power P= VI, with substitution with Ohm,s law also, P=V2/R=I2R, Energy E=Pt. • The power used by the entire circuit P=VI= (88V)(8A)= 704 W o I had the choice of also using V2/R or I2R • The energy used E=Pt= (704W)[(2.16x104 s) =1.53x107J • Using a similar method for the 2 resistor P=56.9 W, and E= 1.23x106J • •