Block A Circuit Theory：块电路理论[精品].ppt

Block A Unit 1 * Parallel network: Proof R1 R2 Is I1 I2 Is = VXY(1/R1 + 1/R2) Replace the parallel network of resistors with a single equivalent resistor RP Is Remember that the voltage across RP is still VXY! VXY/RP = VXY(1/R1 + 1/R2) 1/RP = 1/R1 + 1/R2 X Y Block A Unit 1 * Parallel network (seeing it) R1 R2 Is RN I1 I2 IN Current splits at one node Current re-combine at the other node Suggestion: Think about parallel resistors as the rungs on a ladder Block A Unit 1 * Current divider: Example 1 i1 V2 R1 i2 R2 V1 I Find i1 and i2 in terms of I If R1 = R2, find the ratio between i1 and i2 If R2 = 3R1, find the ratio between i1 and i2 i1 = [R2 / (R1 + R2)]I; i2 = [R1 / (R1 + R2)]I If R1 = R2, then i1 = i2 = I/2 If R2 = 3R1, then i1 = 3I/4, i2 = I/4 Therefore i1 = 3i2 Block A Unit 1 * Current divider: Example 2 Find the current through each resistor in terms of I Find the current through each resistor in terms of I if a 4th resistor was added in parallel How many resistor are required to reduce the current in each resistor to 1% of I (ie 1/100) i1 V2 R i2 R V1 I i3 R i1 = i2 = i3 = I/3 i1 = i2 = i3 = i4 = I/4 Consider that, in = I/n Hence n = 100 for in = 0.01I Block A Unit 1 * Current divider: Example 3 If R2 R4 R1 R5 R3 Which current is the largest? Which current is the smallest? Rank the currents from largest to smallest i1 V2 R1 i2 R2 V1 I i3 R3 i5 R5 i4 R4 I3 I5 I1 I4 I2 Block A Unit 1 * Series network (Highlights) R1 R2 Vs +- RS Vs +- Equivalent Resistance RS = R1 + R2 + …+ RN RN Voltage divider rule VN = VS(RN/RS) + - V1 + - V2 + - VN Block A Unit 1 * Series network: Proof R1 R2 Vs +- + - V1 + - V2 Is Apply Ohm’s law to both resistors: V1 = ISR1; V2 = ISR2 Adding up V1 and V2 according to KVL: Vs = IS(R1+R2) We can now find how much of Vs is distributed between the 2 resistors This is referred to as the voltage divider rule Block A Unit 1 * Series network: Proof R1 R2 Vs +- Vs = IS(R1+R2) RS Vs +- Replace the series network of resistors with a single eq