1 Exam 1 Review Chapters 1, 2, 9. 2 Charge, q Recall Coulomb’s Law Unit: Newton meter 2 / coulomb 2 volt meter / coulomb Charge on an electron (proton)
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2 Charge, q Recall Coulomb’s Law Unit: Newton meter 2 / coulomb 2 volt meter / coulomb Charge on an electron (proton) is negative (positive) and equal to 1.602 x 10 -19 C Force F 1 on charge q 2 due to charge q 1 is given by Note: Positive force is repulsive, negative force is attractive
3 Electric Current, i Current (in amperes) (A) is the time rate of change of charge q 1 A = 1 C/s Charge flowing past a point in the interval [t 0, t] is Convention: Direction of current flow is that of positive charges, opposite to the direction of electron flow
4 Voltage The energy in joules (w) required to move a charge (q) of one coulomb through an element is 1 volt (V). 1 volt = 1 joule/coulomb = 1 newton meter/coulomb
5 Power and Energy Power (p), in watts (W), is the time rate of expending or absorbing energy (w) in joules
6 Power and Energy Change in energy from time t 1 to time t 2 Passive sign convention: + - If p > 0 power is absorbed by the element If p < 0 power is supplied by the element
7 Ohm's Law Units of resistance, R, is Ohms ( ) R = 0: short circuit open circuit
8 Unit of G is siemens (S), Conductance, G 1 S = 1 A/V
9 Power A resistor always dissipates energy; it transforms electrical energy, and dissipates it in the form of heat. Rate of energy dissipation is the instantaneous power
10 Elements in Series Two or more elements are connected in series if they carry the same current and are connected sequentially.
11 Elements in Parallel Two or more elements are connected in parallel if they are connected to the same two nodes & consequently have the same voltage across them.
12 Kirchoff’s Current Law (KCL) The algebraic sum of the currents entering a node (or a closed boundary) is zero. where N = the number of branches connected to the node and i n = the n th current entering (leaving) the node.
13 Sign convention: Currents entering the node are positive, currents leaving the node are negative.
14 Kirchoff’s Current Law (KCL) The algebraic sum of the currents entering (or leaving) a node is zero. The sum of the currents entering a node is equal to the sum of the currents leaving a node. Entering: Leaving:
15 Kirchoff’s Voltage Law (KVL) The algebraic sum of the voltages around any loop is zero. where M = the number of voltages in the loop and v m = the m th voltage in the loop.
16 Sign convention: The sign of each voltage is the polarity of the terminal first encountered in traveling around the loop. The direction of travel is arbitrary. Clockwise: Counter-clockwise:
20 Current Division Current divides in inverse proportion to the resistances
21 Current Division N resistors in parallel Current in j th branch is
22 Source Exchange We can always replace a voltage source in series with a resistor by a current source in parallel with the same resistor and vice-versa. Doing this, however, makes it impossible to directly find the original source current.
25 Writing the Nodal Equations by Inspection The matrix G is symmetric, g kj = g jk and all of the off-diagonal terms are negative or zero. The i k (the k th component of the vector i) = the algebraic sum of the independent currents connected to node k, with currents entering the node taken as positive. The g kj terms are the negative sum of the conductances connected to BOTH node k and node j. The g kk terms are the sum of all conductances connected to node k.
26 The matrix R is symmetric, r kj = r jk and all of the off-diagonal terms are negative or zero. Writing the Mesh Equations by Inspection The v k (the k th component of the vector v) = the algebraic sum of the independent voltages in mesh k, with voltage rises taken as positive. The r kj terms are the negative sum of the resistances common to BOTH mesh k and mesh j. The r kk terms are the sum of all resistances in mesh k.
27 Turning sources off Current source: We replace it by a current source where An open-circuit Voltage source: We replace it by a voltage source where An short-circuit i
28 Thevenin's Theorem Thevenin’s theorem states that the two circuits given below are equivalent as seen from the load R L that is the same in both cases. V Th = Thevenin’s voltage = V ab with R L disconnected (= ) = the open-circuit voltage = V OC
29 Thevenin's Theorem R Th = Thevenin’s resistance = the input resistance with all independent sources turned off (voltage sources replaced by short circuits and current sources replaced by open circuits). This is the resistance seen at the terminals ab when all independent sources are turned off.
31 Maximum Power Transfer What is the resistance of the load that will result in the maximum power being delivered to the load? Consider the source to be modeled by its Thevenin equivalent.
32 Thus, maximum power transfer takes place when the resistance of the load equals the Thevenin resistance R Th. Note also that Thus, at best, one-half of the power is dissipated in the internal resistance and one-half in the load.
33 Ideal Op Amp 1) The open-loop gain, A v, is very large, approaching infinity. 2) The current into the inputs are zero.
34 Ideal Op Amp with Negative Feedback Golden Rules of Op Amps: 1.The output attempts to do whatever is necessary to make the voltage difference between the inputs zero. 2.The inputs draw no current.
35 Non-inverting Amplifier Closed-loop voltage gain
36 Inverting Amplifier Current into op amp is zero