EEE107J1

Internal Resistance of a Voltage Source Ideal voltage source – no internal resistance Rint = internal resistance of source No Load Voltage Load Voltage after Boylestad

Find VL and power loss in Rint after Boylestad

Voltage Regulation. after Boylestad

Identical resistors in parallel If all the resistances have equal value then after Boylestad

Four parallel resistors of equal value. after Boylestad

Calculate RT, IS, I1,I2, Power in each resistor and total power dissipated after Boylestad

Determine R3, E, Is, I2, after Boylestad

Using Kirchoff’s current law Find I5 after Boylestad

Find I3 and I7 after Boylestad

Current division. For two parallel elements of equal value the current will divide equally For parallel elements with different values the smaller the resistance, the greater the share of input current. For parallel elements of different values, the current will split with a ratio equal to the inverse of their resistor values after Boylestad

Current divider rule for two resistors in parallel after Boylestad

Calculate the current through each resistor after Boylestad

Determine I1 and I2 I2 could be found by the application of the current rule after Boylestad

Determine R1 Use KCL Now use Ohm’s law to find VR2 VR2 = VR1 i.e. Ohm’s law

Demonstrating the characteristics of an open circuit. after Boylestad

Demonstrating the effect of a short circuit on current levels. after Boylestad

Basically follow the following steps Branch current analysis This is a means of analysing a network by the application of Kirchhoff’s laws to linear networks Basically follow the following steps Assign a distinct current of arbitrary direction to each branch of the network. Indicate the polarities for each resistor as determined by the assumed current directions Apply Kirchhoff’s voltage law around each closed independent loop of the network

B A C F E D Assign currents I1, I2, and I3 Note I1+I2=I3 Assign letters/identification to network

A B C F D E Insert the polarities across the resistive elements as defined by the chosen branch currents.

Apply KVL to each loop ABEF 2 = 2I1 + 4I3 BCDE 6 = 1I2 + 4I3 OR 2 = 2I1 + 4I1 + 4I2 OR 6 = 1I2 + 4I1 + 4I2

Combining terms in this set of equations gives ABEF 2 = 2I1 + 4I3 BCDE 6 = 1I2 + 4I3 OR 2 = 2I1 + 4I1 + 4I2 OR 6 = 1I2 + 4I1 + 4I2 It should be clear that the equations on the right have only two unknown currents I1 and I2. Combining terms in this set of equations gives ABEF 2 = 2I1 + 4I1 + 4I2 gives 2 = 6I1+4I2 BCDE 6 = 1I2 + 4I1 + 4I2 gives 6 = 4I1+5I2

ABEF 2 = 6I1+ 4I2 BCDE 6 = 4I1+ 5I2 multiplying ABEF by 2. gives multiplying BCDE by 3 gives ABEF 4 = 12I1+ 8I2 BCDE 18 = 12I1+15I2 Subtracting BCDE from ABEF gives -14 = -7I2 thus I2 = 2A Substituting I2 = 2A back into the equation for loop ABEF 2=6I1+8 ABEF 2 =6I1+4I2 Thus I1=-1A note the current is minus and thus flows in the opposite direction to that originally assigned

Reviewing the results of the analysis of the network

Calculate V1 and V2 assume that IB = 0A Determine the voltage VE and current IE Determine VRC assuming IC = IE Calculate VCE 25 25

Calculate V1 and V2 assume that IB = 0A Apply voltage divider rule to R1 R2 Apply KVL to find V1 22V = V1 + V2 = V1 + 2V V1 = 20V after Boylestad 26 26

Determine the voltage VE and current IE V1 = 20V , V2 = 2V Determine the voltage VE and current IE Apply KVL to base circuit as below 27 27

Determine the voltage VE and current IE V1 = 20V , V2 = 2V Determine the voltage VE and current IE Apply KVL to base circuit as below In this loop V2 + VBE + VE = 0V Apply potentials and polarities after Boylestad 28 28

Determine VRC assuming IC = IE V1 = 20V , V2 = 2V, VE = 1.3V, IE = 1.3mA. Calculate VRC Determine VRC assuming IC = IE after Boylestad 29 29