Exam 1 Review Chapters 1, 2, 9

Charge, q Recall Coulomb’s Law Force F1 on charge q2 due to charge q1 is given by Charge on an electron (proton) is negative (positive) and equal to 1.602 x 10-19 C Unit: Newton meter2 / coulomb2 volt meter / coulomb Note: Positive force is repulsive, negative force is attractive

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 [t0, t] is Convention: Direction of current flow is that of positive charges, opposite to the direction of electron flow

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

Power and Energy Power (p), in watts (W), is the time rate of expending or absorbing energy (w) in joules

Power and Energy Change in energy from time t1 to time t2 Passive sign convention: If p > 0 power is absorbed by the element If p < 0 power is supplied by the element + -

Ohm's Law Units of resistance, R, is Ohms (W) R = 0: short circuit open circuit

Conductance, G Unit of G is siemens (S), 1 S = 1 A/V

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

Elements in Series Two or more elements are connected in series if they carry the same current and are connected sequentially.

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.

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 in = the nth current entering (leaving) the node.

Sign convention: Currents entering the node are positive, currents leaving the node are negative.

Kirchoff’s Current Law (KCL) The algebraic sum of the currents entering (or leaving) a node is zero. Entering: Leaving: The sum of the currents entering a node is equal to the sum of the currents leaving a node.

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 vm = the mth voltage in the loop.

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:

Series Resistors

Voltage Divider

Parallel Resistors

Current Division Current divides in inverse proportion to the resistances

Current Division N resistors in parallel Current in jth branch is

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.

Source Exchange Proof Voltage across and current through any load are the same

3-bit R2-R Ladder Network KCL

KCL

KCL

Writing the Nodal Equations by Inspection The matrix G is symmetric, gkj = gjk and all of the off-diagonal terms are negative or zero. The gkk terms are the sum of all conductances connected to node k. The gkj terms are the negative sum of the conductances connected to BOTH node k and node j. The ik (the kth component of the vector i) = the algebraic sum of the independent currents connected to node k, with currents entering the node taken as positive.

Writing the Mesh Equations by Inspection The matrix R is symmetric, rkj = rjk and all of the off-diagonal terms are negative or zero. The rkk terms are the sum of all resistances in mesh k. The rkj terms are the negative sum of the resistances common to BOTH mesh k and mesh j. The vk (the kth component of the vector v) = the algebraic sum of the independent voltages in mesh k, with voltage rises taken as positive.

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 i An short-circuit

Thevenin's Theorem Thevenin’s theorem states that the two circuits given below are equivalent as seen from the load RL that is the same in both cases. VTh = Thevenin’s voltage = Vab with RL disconnected (= ) = the open-circuit voltage = VOC

Thevenin's Theorem RTh = 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.

Example

Maximum Power Transfer In all practical cases, energy sources have non-zero internal resistance. Thus, there are losses inherent in any real source. Also, in most cases the aim of an energy source is to provide power to a load. Given a circuit with a known internal resistance, 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.

The power delivered to the load (absorbed by RL) is This power is maximum when

Thus, maximum power transfer takes place when the resistance of the load equals the Thevenin resistance RTh. Note also that Thus, at best, one-half of the power is dissipated in the internal resistance and one-half in the load.

Ideal Op Amp 1) The open-loop gain, Av, is very large, approaching infinity. 2) The current into the inputs are zero.

Ideal Op Amp with Negative Feedback Golden Rules of Op Amps: The output attempts to do whatever is necessary to make the voltage difference between the inputs zero. The inputs draw no current.

Non-inverting Amplifier Closed-loop voltage gain

Inverting Amplifier Current into op amp is zero