EE201 Lecture 3 P 1 Resistance Ohm s Law Power Resistance Resistance R is a measure of whether current flow is easy or difficult A resistor is a material that is designed to have a specific resistance Note Resistance depends on size and shape of resistor For a wire R as length and diam R L A resistivity Unit of resistance is Ohm I V Symbol for a resistor with passive sign labeling EE201 Lecture 3 P 2 I V From Ohm s observations Proportionality constant is resistance R Ohm s Law V RI or I 1 R V G V where G semens mhos conductance is the A material has a 1 resistance if a one volt excitation causes one ampere of current to flow EE201 Lecture 3 P 3 Implications of resistance on power calculations p t v t i t Substitute for v t p t R i t i t R i2 t Substitute for i t p t v t v t R v2 t R For dc conditions P V I I2 R V2 R EE201 Lecture 3 Example P 4 I 24 5 W 120 V 120 V 18 5 W V1 R1 V2 R2 Find a the current i t through each bulb b the voltages V1 t and V2 t across each bulb c the resistance of each bulb d the power used by each bulb when connected directly to the 120 V source Note Resistors always absorb power EE201 Lecture 3 P 5 Solution a Total power 43 W P V I 120 V I 43 W I 0 358 A b From definition of power V P I V1 P1 I 24 5 W 0 358 A 68 4 V V2 P2 I 18 5 W 0 358 A 51 6 V c From Ohm s Law R V I R1 V1 I 68 4 V 0 358 A 190 8 R2 V2 I 51 6 V 0 358 A 144 0 d P1 V2 R1 120 V 2 190 8 75 5 W P2 V2 R2 120 V 2 144 0 100 W EE201 Lecture 3 P 6 Example 85 V iin t 10 W 15 W 20 W 25 W Find a the resistance of each bulb b the total power delivered by the source and c if at t 10 h the current supplied by the source is 0 647 A discover which bulb has burnt out EE201 Lecture 3 P 7 Solution a 10 W R V2 P 85 V 2 10 W 722 5 15 W R V2 P 85 V 2 15 W 481 6 20 W R V2 P 85 V 2 20 W 361 3 25 W R V2 P 85 V 2 25 W 289 0 b From principle of conservation of power Total power 70 W c source current I 70 W 85 V 0 824 A 10 W I P V 10 W 85 V 0 118 15 W I P V 15 W 85 V 0 176 A 20 W I P V 20 W 85 V 0 235 A 25 W I P V 25 W 85 V 0 294 A If current drawn is 0 647 A after bulb is burnt out and 0 824 A before burn out then current of burned out bulb 0 823 0 647 0 176 A EE201 Lecture 3 P 8 Simple circuits all two terminal elements carry the same current Reason we assume circuits are composed of lumped elements Lumped elements have dimensions smaller than the wavelength of the excitation If current is i t A sin t the frequency is 2 The wavelength is c where c is the velocity of light 3 0 x 108 m s 1 5 1 0 0 5 0 0 0 5 1 0 1 5 i t x Current does not change much if element is much smaller than EE201 Lecture 3 P 9 Dependent power sources denoted by diamond i t f vx i t f ix VCCS v t f vx VCVS CCCS v t f ix CCVS x denotes an element other than the source
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